To the Student As you begin, you may feel anxious about the number of theorems, definitions, procedures, and equations. You may wonder if you can learn it all in time. Don’t worry—your concerns are normal. This textbook was written with you in mind. If you attend class, work hard, and read and study this text, you will build the knowledge and skills you need to be successful. Here’s how you can use the text to your benefit. Read Carefully When you get busy, it’s easy to skip reading and go right to the problems. Don’t . . . the text has a large number of examples and clear explanations to help you break down the mathematics into easy-to-understand steps. Reading will provide you with a clearer understanding, beyond simple memorization. Read before class (not after) so you can ask questions about anything you didn’t understand. You’ll be amazed at how much more you’ll get out of class if you do this. Use the Features I use many different methods in the classroom to communicate. Those methods, when incorporated into the text, are called “features.” The features serve many purposes, from providing timely review of material you learned before (just when you need it) to providing organized review sessions to help you prepare for quizzes and tests. Take advantage of the features and you will master the material. To make this easier, we’ve provided a brief guide to getting the most from this text. Refer to “Prepare for Class,” “Practice,” and “Review” on the following three pages. Spend fifteen minutes reviewing the guide and familiarizing yourself with the features by flipping to the page numbers provided. Then, as you read, use them. This is the best way to make the most of your text. Please do not hesitate to contact us, through Pearson Education, with any questions, comments, or suggestions for improving this text. I look forward to hearing from you, and good luck with all of your studies. Best Wishes! Feature Description Benefit Page Every Chapter Opener begins with . . . Chapter- Opening Each chapter begins with a discussion of a topic of current interest and ends with a Topic & Project The Project lets you apply what you learned to solve a problem related to the topic. 402 The projects allow for the integration of spreadsheet technology that you will need to be a productive member of the workforce. The projects give you an opportunity to collaborate and use mathematics to deal with issues of current interest. 503 Each section begins with a list of objectives. Objectives also appear in the text where the objective is covered. These focus your studying by emphasizing what’s most important and where to find it. 423 PREPARING FOR THIS SECTION Most sections begin with a list of key concepts to review with page numbers. Ever forget what you’ve learned? This feature highlights previously learned material to be used in this section. Review it, and you’ll always be prepared to move forward. 423 Now Work the ‘Are You Prepared?’ Problems Problems that assess whether you have the Not sure you need the Preparing for This 423, 434 prerequisite knowledge for the upcoming Section review? Work the ‘Are You section. Prepared?’ problems. If you get one wrong, you’ll know exactly what you need to review and where to review it! Now Work These follow most examples and direct you to a related exercise. related project. Internet-Based Projects Every Section begins with . . . Learning Objectives 2 Sections contain . . . PROBLEMS WARNING Warnings are provided in the text. We learn best by doing. You’ll solidify your understanding of examples if you try a similar problem right away, to be sure you understand what you’ve just read. These point out common mistakes and help you to avoid them. 430, 435 456 418, 443 These graphing utility activities foreshadow a concept or solidify a concept just presented. You will obtain a deeper and more intuitive understanding of theorems and definitions. In Words These provide alternative descriptions of select definitions and theorems. Does math ever look foreign to you? This feature translates math into plain English. Calculus These appear next to information essential for the study of calculus. Pay attention–if you spend extra time now, you’ll do better later! These examples provide “how-to” instruction by offering a guided, step-by-step approach to solving a problem. With each step presented on the left and the mathematics displayed on the right, you can immediately see how each step is employed. 334 These examples and problems require you to build a mathematical model from either a verbal description or data. The homework Model It! problems are marked by purple headings. It is rare for a problem to come in the form “Solve the following equation.” Rather, the equation must be developed based on an explanation of the problem. These problems require you to develop models that will allow you to describe the problem mathematically and suggest a solution to the problem. 447, 475 Exploration and Seeing the Concept SHOWCASE EXAMPLES Model It! Examples and Problems 440 205, 407, 431 Feature Description Benefit Page ‘Are You Prepared?’ Problems These assess your retention of the prerequisite material you’ll need. Answers are given at the end of the section exercises. This feature is related to the Preparing for This Section feature. Do you always remember what you’ve learned? Working these problems is the best way to find out. If you get one wrong, you’ll know exactly what you need to review and where to review it! 434, 440 Concepts and Vocabulary These short-answer questions, mainly Fill-in-the-Blank, Multiple-Choice and True/False items, assess your understanding of key definitions and concepts in the current section. It is difficult to learn math without knowing the language of mathematics. These problems test your understanding of the formulas and vocabulary. 434 Skill Building Correlated with section examples, these problems provide straightforward practice. It’s important to dig in and develop your skills. These problems provide you with ample opportunity to do so. 434–436 Mixed Practice These problems offer comprehensive assessment of the skills learned in the section by asking problems that relate to more than one concept or objective. These problems may also require you to utilize skills learned in previous sections. Learning mathematics is a building process. Many concepts are interrelated. These problems help you see how mathematics builds on itself and also see how the concepts tie together. 436–437 Applications and Extensions These problems allow you to apply your skills to real-world problems. They also allow you to extend concepts learned in the section. You will see that the material learned within the section has many uses in everyday life. 437–439 Explaining Concepts: “Discussion and Writing” problems are colored red. They support class Discussion and discussion, verbalization of mathematical Writing To verbalize an idea, or to describe it clearly in writing, shows real understanding. These problems nurture that understanding. Many are challenging, but you’ll get out what you put in. 439 NEW! Retain Your Knowledge These problems allow you to practice content learned earlier in the course. Remembering how to solve all the different kinds of problems that you encounter throughout the course is difficult. This practice helps you remember. 439 Now Work Many examples refer you to a related homework problem. These related problems are marked by a pencil and orange numbers. If you get stuck while working problems, look for the closest Now Work problem, and refer to the related example to see if it helps. Every chapter concludes with a comprehensive list of exercises to pratice. Use the list of objectives to determine the objective and examples that correspond to the problems. Work these problems to ensure that you 499–501 understand all the skills and concepts of the chapter. Think of it as a comprehensive review of the chapter. ideas, and writing and research projects. PROBLEMS Review Exercises 432, 435, 436 Feature Description Benefit Page The Chapter Review at the end of each chapter contains . . . Things to Know A detailed list of important theorems, formulas, and definitions from the chapter. Review these and you’ll know the most important material in the chapter! You Should Be Able to . . . Contains a complete list of objectives by section, examples that illustrate the objective, and practice exercises that test your understanding of the objective. Do the recommended exercises and you’ll 498–499 have mastered the key material. If you get something wrong, go back and work through the example listed and try again. Review Exercises These provide comprehensive review and Practice makes perfect. These problems 499–501 practice of key skills, matched to the Learning combine exercises from all sections, giving you a comprehensive review in one Objectives for each section. place. Chapter Test About 15–20 problems that can be taken Be prepared. Take the sample practice as a Chapter Test. Be sure to take the Chapter test under test conditions. This will get you ready for your instructor’s test. If you get a Test under test conditions—no notes! problem wrong, you can watch the Chapter Test Prep Video. Cumulative Review These problem sets appear at the end of each chapter, beginning with Chapter 2. They combine problems from previous chapters, providing an ongoing cumulative review. When you use them in conjunction with the Retain Your Knowledge problems, you will be ready for the final exam. These problem sets are really important. 502–503 Completing them will ensure that you are not forgetting anything as you go. This will go a long way toward keeping you primed for the final exam. Chapter Projects The Chapter Projects apply to what you’ve learned in the chapter. Additional projects are available on the Instructor’s Resource Center (IRC). The Chapter Projects give you an opportunity 503–504 to apply what you’ve learned in the chapter to the opening topic. If your instructor allows, these make excellent opportunities to work in a group, which is often the best way of learning math. Internet-Based Projects In selected chapters, a Web-based project These projects give you an opportunity to is given. collaborate and use mathematics to deal with issues of current interest by using the Internet to research and collect data. 497–498 502 503 Achieve Your Potential 7KHDXWKRU0LFKDHO6XOOLYDQKDVGHYHORSHGVSHFLÀFFRQWHQW in MyMathLab® to ensure you have many resources to help you achieve success in mathematics - and beyond! The MyMathLab features described here will help you: Review math skills and concepts you may have forgotten Retain new concepts as you move through your math course Develop skills that will help with your transition to college Adaptive Study Plan The Study Plan will help you study PRUHHIÀFLHQWO\DQGHIIHFWLYHO\ Your performance and activity are assessed continually in real time, providing a personalized experience EDVHGRQ\RXULQGLYLGXDOQHHGV Skills for Success The Skills for Success Modules support your continued success in FROOHJHThese modules provide tutorials and guidance on a variety of topics, including transitioning to college, online learning, time PDQDJHPHQWDQGPRUH Additional content is provided to help with the development of professional skills such as resume ZULWLQJDQGLQWHUYLHZSUHSDUDWLRQ Getting Ready Are you frustrated when you know you learned a math concept in the past, but you can’t quite remember the skill when it’s time to use it? Don’t worry! The author has included Getting Ready material so you can brush up on forgotten PDWHULDOHIÀFLHQWO\E\WDNLQJD TXLFNVNLOOUHYLHZTXL]WRSLQSRLQWWKHDUHDVZKHUH\RXQHHGKHOS Then, a personalized homework assignment provides additional SUDFWLFHRQWKRVHIRUJRWWHQFRQFHSWVULJKWZKHQ\RXQHHGLW Retain Your Knowledge As you work through your math course, these MyMathLab® exercises support ongoing review to help you maintain HVVHQWLDOVNLOOV The ability to recall important math concepts as you continually acquire new mathematical skills will help you be successful in this math course and in your IXWXUHPDWKFRXUVHV Algebra & Trigonometry Tenth Edition Michael Sullivan Chicago State University Boston Columbus Indianapolis New York San Francisco Hoboken Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Editor in Chief: Anne Kelly Acquisitions Editor: Dawn Murrin Assistant Editor: Joseph Colella Program Team Lead: Marianne Stepanian Program Manager: Chere Bemelmans Project Team Lead: Peter Silvia Project Manager: Peggy McMahon Associate Media Producer: Marielle Guiney Senior Project Manager, MyMathLab: Kristina Evans QA Manager, Assessment Content: Marty Wright Senior Marketing Manager: Michelle Cook Marketing Manager: Peggy Sue Lucas Marketing Assistant: Justine Goulart Senior Author Support/Technology Specialist: Joe Vetere Procurement Manager: Vincent Scelta Procurement Specialist: Carol Melville Text Design: Tamara Newnam Production Coordination, Composition, Illustrations: Cenveo® Publisher Services Associate Director of Design, USHE EMSS/HSC/EDU: Andrea Nix Project Manager, Rights and Permissions: Diahanne Lucas Dowridge Art Director: Heather Scott Cover Design and Cover Illustration: Tamara Newnam Acknowledgments of third-party content appear on page C1, which constitutes an extension of this copyright page. Unless otherwise indicated herein, any third-party trademarks that may appear in this work are the property of their respective owners, and any references to third-party trademarks, logos or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson’s products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc. or its affiliates, authors, licensees or distributors. MICROSOFT® AND WINDOWS® ARE REGISTERED TRADEMARKS OF THE MICROSOFT CORPORATION IN THE U.S.A. AND OTHER COUNTRIES. SCREEN SHOTS AND ICONS REPRINTED WITH PERMISSION FROM THE MICROSOFT CORPORATION. THIS BOOK IS NOT SPONSORED OR ENDORSED BY OR AFFILIATED WITH THE MICROSOFT CORPORATION. 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THE DOCUMENTS AND RELATED GRAPHICS CONTAINED HEREIN COULD INCLUDE TECHNICAL INACCURACIES OR TYPOGRAPHICAL ERRORS. CHANGES ARE PERIODICALLY ADDED TO THE INFORMATION HEREIN. MICROSOFT AND/OR ITS RESPECTIVE SUPPLIERS MAY MAKE IMPROVEMENTS AND /OR CHANGES IN THE PRODUCT (S) AND /OR THE PROGRAM (S) DESCRIBED HEREIN AT ANY TIME. PARTIAL SCREEN SHOTS MAY BE VIEWED IN FULL WITHIN THE SOFTWARE VERSION SPECIFIED. The student edition of this text has been cataloged as follows: Library of Congress Cataloging-in-Publication Data Sullivan, Michael, 1942Algebra & trigonometry / Michael Sullivan, Chicago State University. -- Tenth edition. pages cm. ISBN 978-0-321-99859-0 1. Algebra--Textbooks. 2. Algebra--Study and teaching (Higher) 3. Trigonometry--Textbooks. 4. Trigonometry--Study and teaching (Higher) I. Title. II. Title: Algebra and trigonometry. QA154.3.S73 2016 512’ .13--dc23 2014021731 Copyright © 2016 by Pearson Education, Inc. or its affiliates. All Rights Reserved. Printed in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms, and the appropriate contacts within the Pearson Education Global Rights & Permissions department, please visit www.pearsoned.com/permissions/. PEARSON, ALWAYS LEARNING, and MYMATHLAB are exclusive trademarks in the U.S. and/or other countries owned by Pearson Education, Inc. or its affiliates. 1 2 3 4 5 6 7 8 9 10—CRK—17 16 15 14 www.pearsonhighered.com ISBN-10: 0-321-99859-6 ISBN-13: 978-0-321-99859-0 Contents Three Distinct Series R xviii The Contemporary Series xix Preface to the Instructor xx Resources for Success xxiv Applications Index xxvi Review 1 R.1 Real Numbers 2 8PSLXJUI4FUTr$MBTTJGZ/VNCFSTr&WBMVBUF/VNFSJDBM&YQSFTTJPOT r8PSLXJUI1SPQFSUJFTPG3FBM/VNCFST R.2 Algebra Essentials 17 (SBQI*OFRVBMJUJFTr'JOE%JTUBODFPOUIF3FBM/VNCFS-JOFr&WBMVBUF "MHFCSBJD&YQSFTTJPOTr%FUFSNJOFUIF%PNBJOPGB7BSJBCMFr6TFUIF -BXTPG&YQPOFOUTr&WBMVBUF4RVBSF3PPUTr6TFB$BMDVMBUPSUP&WBMVBUF &YQPOFOUTr6TF4DJFOUJGJD/PUBUJPO R.3 Geometry Essentials 30 6TFUIF1ZUIBHPSFBO5IFPSFNBOE*UT$POWFSTFr,OPX(FPNFUSZ 'PSNVMBTr6OEFSTUBOE$POHSVFOU5SJBOHMFTBOE4JNJMBS5SJBOHMFT R.4 Polynomials 39 3FDPHOJ[F.POPNJBMTr3FDPHOJ[F1PMZOPNJBMTr"EEBOE4VCUSBDU 1PMZOPNJBMTr.VMUJQMZ1PMZOPNJBMTr,OPX'PSNVMBTGPS4QFDJBM1SPEVDUT r%JWJEF1PMZOPNJBMT6TJOH-POH%JWJTJPOr8PSLXJUI1PMZOPNJBMTJO5XP Variables R.5 Factoring Polynomials 49 Factor the Difference of Two Squares and the Sum and Difference of Two $VCFTr'BDUPS1FSGFDU4RVBSFTr'BDUPSB4FDPOE%FHSFF Polynomial: x2 + Bx + Cr'BDUPSCZ(SPVQJOHr'BDUPSB4FDPOE%FHSFF Polynomial: Ax2 + Bx + C, A ≠ 1r$PNQMFUFUIF4RVBSF R.6 Synthetic Division 58 Divide Polynomials Using Synthetic Division R.7 Rational Expressions 62 3FEVDFB3BUJPOBM&YQSFTTJPOUP-PXFTU5FSNTr.VMUJQMZBOE%JWJEF 3BUJPOBM&YQSFTTJPOTr"EEBOE4VCUSBDU3BUJPOBM&YQSFTTJPOTr6TFUIF -FBTU$PNNPO.VMUJQMF.FUIPEr4JNQMJGZ$PNQMFY3BUJPOBM&YQSFTTJPOT R.8 nth Roots; Rational Exponents 73 Work with nUI3PPUTr4JNQMJGZ3BEJDBMTr3BUJPOBMJ[F%FOPNJOBUPST r4JNQMJGZ&YQSFTTJPOTXJUI3BUJPOBM&YQPOFOUT 1 Equations and Inequalities 1.1 Linear Equations 81 82 4PMWFB-JOFBS&RVBUJPOr4PMWF&RVBUJPOT5IBU-FBEUP-JOFBS&RVBUJPOT r4PMWF1SPCMFNT5IBU$BO#F.PEFMFECZ-JOFBS&RVBUJPOT 1.2 Quadratic Equations 92 4PMWFB2VBESBUJD&RVBUJPOCZ'BDUPSJOHr4PMWFB2VBESBUJD&RVBUJPOCZ $PNQMFUJOHUIF4RVBSFr4PMWFB2VBESBUJD&RVBUJPO6TJOHUIF2VBESBUJD 'PSNVMBr4PMWF1SPCMFNT5IBU$BO#F.PEFMFECZ2VBESBUJD&RVBUJPOT vii viii CONTENTS 1.3 Complex Numbers; Quadratic Equations in the Complex Number System 104 Add, Subtract, Multiply, and Divide Complex Numbers r4PMWF2VBESBUJD&RVBUJPOTJOUIF$PNQMFY/VNCFS4ZTUFN 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations 113 4PMWF3BEJDBM&RVBUJPOTr4PMWF&RVBUJPOT2VBESBUJDJO'PSNr4PMWF Equations by Factoring 1.5 Solving Inequalities 119 6TF*OUFSWBM/PUBUJPOr6TF1SPQFSUJFTPG*OFRVBMJUJFTr4PMWF *OFRVBMJUJFTr4PMWF$PNCJOFE*OFRVBMJUJFT 1.6 Equations and Inequalities Involving Absolute Value 130 4PMWF&RVBUJPOT*OWPMWJOH"CTPMVUF7BMVFr4PMWF*OFRVBMJUJFT*OWPMWJOH Absolute Value 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 134 5SBOTMBUF7FSCBM%FTDSJQUJPOTJOUP.BUIFNBUJDBM&YQSFTTJPOTr4PMWF *OUFSFTU1SPCMFNTr4PMWF.JYUVSF1SPCMFNTr4PMWF6OJGPSN.PUJPO 1SPCMFNTr4PMWF$POTUBOU3BUF+PC1SPCMFNT 2 Chapter Review 143 Chapter Test 147 Chapter Projects 147 Graphs 2.1 The Distance and Midpoint Formulas 149 150 6TFUIF%JTUBODF'PSNVMBr6TFUIF.JEQPJOU'PSNVMB 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry 157 (SBQI&RVBUJPOTCZ1MPUUJOH1PJOUTr'JOE*OUFSDFQUTGSPNB(SBQIr'JOE *OUFSDFQUTGSPNBO&RVBUJPOr5FTUBO&RVBUJPOGPS4ZNNFUSZXJUI3FTQFDU to the x-Axis, the y"YJT BOEUIF0SJHJOr,OPX)PXUP(SBQI,FZ Equations 2.3 Lines 167 $BMDVMBUFBOE*OUFSQSFUUIF4MPQFPGB-JOFr(SBQI-JOFT(JWFOB1PJOU BOEUIF4MPQFr'JOEUIF&RVBUJPOPGB7FSUJDBM-JOFr6TFUIF1PJOUm4MPQF 'PSNPGB-JOF*EFOUJGZ)PSJ[POUBM-JOFTr'JOEUIF&RVBUJPOPGB-JOF (JWFO5XP1PJOUTr8SJUFUIF&RVBUJPOPGB-JOFJO4MPQFm*OUFSDFQU'PSN r*EFOUJGZUIF4MPQFBOEy*OUFSDFQUPGB-JOFGSPN*UT&RVBUJPOr(SBQI -JOFT8SJUUFOJO(FOFSBM'PSN6TJOH*OUFSDFQUTr'JOE&RVBUJPOTPG1BSBMMFM -JOFTr'JOE&RVBUJPOTPG1FSQFOEJDVMBS-JOFT 2.4 Circles 182 8SJUFUIF4UBOEBSE'PSNPGUIF&RVBUJPOPGB$JSDMFr(SBQIB$JSDMF r8PSLXJUIUIF(FOFSBM'PSNPGUIF&RVBUJPOPGB$JSDMF 2.5 Variation 188 $POTUSVDUB.PEFM6TJOH%JSFDU7BSJBUJPOr$POTUSVDUB.PEFM6TJOH *OWFSTF7BSJBUJPOr$POTUSVDUB.PEFM6TJOH+PJOU7BSJBUJPOPS$PNCJOFE Variation Chapter Review 194 Chapter Test 196 Cumulative Review 196 Chapter Project 197 CONTENTS 3 Functions and Their Graphs 198 3.1 Functions 199 %FUFSNJOF8IFUIFSB3FMBUJPO3FQSFTFOUTB'VODUJPOr'JOEUIF7BMVFPGB 'VODUJPOr'JOEUIF%JGGFSFODF2VPUJFOUPGB'VODUJPOr'JOEUIF%PNBJO PGB'VODUJPO%FGJOFECZBO&RVBUJPOr'PSNUIF4VN %JGGFSFODF 1SPEVDU and Quotient of Two Functions 3.2 The Graph of a Function 214 *EFOUJGZUIF(SBQIPGB'VODUJPOr0CUBJO*OGPSNBUJPOGSPNPSBCPVUUIF Graph of a Function 3.3 Properties of Functions 223 %FUFSNJOF&WFOBOE0EE'VODUJPOTGSPNB(SBQIr*EFOUJGZ&WFOBOE 0EE'VODUJPOTGSPNBO&RVBUJPOr6TFB(SBQIUP%FUFSNJOF8IFSFB 'VODUJPO*T*ODSFBTJOH %FDSFBTJOH PS$POTUBOUr6TFB(SBQIUP-PDBUF -PDBM.BYJNBBOE-PDBM.JOJNBr6TFB(SBQIUP-PDBUFUIF"CTPMVUF .BYJNVNBOEUIF"CTPMVUF.JOJNVNr6TFB(SBQIJOH6UJMJUZUP Approximate Local Maxima and Local Minima and to Determine Where a 'VODUJPO*T*ODSFBTJOHPS%FDSFBTJOHr'JOEUIF"WFSBHF3BUFPG$IBOHF of a Function 3.4 Library of Functions; Piecewise-defined Functions 237 (SBQIUIF'VODUJPOT-JTUFEJOUIF-JCSBSZPG'VODUJPOTr(SBQI Piecewise-defined Functions 3.5 Graphing Techniques: Transformations 247 (SBQI'VODUJPOT6TJOH7FSUJDBMBOE)PSJ[POUBM4IJGUTr(SBQI'VODUJPOT 6TJOH$PNQSFTTJPOTBOE4USFUDIFTr(SBQI'VODUJPOT6TJOH3FGMFDUJPOT about the x-Axis and the y-Axis 3.6 Mathematical Models: Building Functions 260 Build and Analyze Functions 4 Chapter Review 266 Chapter Test 270 Cumulative Review 271 Chapter Projects 271 Linear and Quadratic Functions 273 4.1 Properties of Linear Functions and Linear Models 274 (SBQI-JOFBS'VODUJPOTr6TF"WFSBHF3BUFPG$IBOHFUP*EFOUJGZ-JOFBS 'VODUJPOTr%FUFSNJOF8IFUIFSB-JOFBS'VODUJPO*T*ODSFBTJOH %FDSFBTJOH PS$POTUBOUr#VJME-JOFBS.PEFMTGSPN7FSCBM%FTDSJQUJPOT 4.2 Building Linear Models from Data 284 %SBXBOE*OUFSQSFU4DBUUFS%JBHSBNTr%JTUJOHVJTICFUXFFO-JOFBS BOE/POMJOFBS3FMBUJPOTr6TFB(SBQIJOH6UJMJUZUP'JOEUIF-JOF of Best Fit 4.3 Quadratic Functions and Their Properties 290 (SBQIB2VBESBUJD'VODUJPO6TJOH5SBOTGPSNBUJPOTr*EFOUJGZUIF7FSUFY BOE"YJTPG4ZNNFUSZPGB2VBESBUJD'VODUJPOr(SBQIB2VBESBUJD 'VODUJPO6TJOH*UT7FSUFY "YJT BOE*OUFSDFQUTr'JOEB2VBESBUJD'VODUJPO (JWFO*UT7FSUFYBOE0OF0UIFS1PJOUr'JOEUIF.BYJNVNPS.JOJNVN Value of a Quadratic Function 4.4 Build Quadratic Models from Verbal Descriptions and from Data 302 #VJME2VBESBUJD.PEFMTGSPN7FSCBM%FTDSJQUJPOTr#VJME2VBESBUJD.PEFMT from Data ix x CONTENTS 4.5 Inequalities Involving Quadratic Functions 312 Solve Inequalities Involving a Quadratic Function 5 Chapter Review 315 Chapter Test 318 Cumulative Review 319 Chapter Projects 320 Polynomial and Rational Functions 5.1 Polynomial Functions and Models 321 322 *EFOUJGZ1PMZOPNJBM'VODUJPOTBOE5IFJS%FHSFFr(SBQI1PMZOPNJBM 'VODUJPOT6TJOH5SBOTGPSNBUJPOTr*EFOUJGZUIF3FBM;FSPTPGB1PMZOPNJBM 'VODUJPOBOE5IFJS.VMUJQMJDJUZr"OBMZ[FUIF(SBQIPGB1PMZOPNJBM 'VODUJPOr#VJME$VCJD.PEFMTGSPN%BUB 5.2 Properties of Rational Functions 343 'JOEUIF%PNBJOPGB3BUJPOBM'VODUJPOr'JOEUIF7FSUJDBM"TZNQUPUFT PGB3BUJPOBM'VODUJPOr'JOEUIF)PSJ[POUBMPS0CMJRVF"TZNQUPUFPGB Rational Function 5.3 The Graph of a Rational Function 353 "OBMZ[FUIF(SBQIPGB3BUJPOBM'VODUJPOr4PMWF"QQMJFE1SPCMFNT Involving Rational Functions 5.4 Polynomial and Rational Inequalities 368 4PMWF1PMZOPNJBM*OFRVBMJUJFTr4PMWF3BUJPOBM*OFRVBMJUJFT 5.5 The Real Zeros of a Polynomial Function 375 6TFUIF3FNBJOEFSBOE'BDUPS5IFPSFNTr6TF%FTDBSUFT3VMFPG4JHOTUP %FUFSNJOFUIF/VNCFSPG1PTJUJWFBOEUIF/VNCFSPG/FHBUJWF3FBM;FSPT PGB1PMZOPNJBM'VODUJPOr6TFUIF3BUJPOBM;FSPT5IFPSFNUP-JTUUIF 1PUFOUJBM3BUJPOBM;FSPTPGB1PMZOPNJBM'VODUJPOr'JOEUIF3FBM;FSPTPG B1PMZOPNJBM'VODUJPOr4PMWF1PMZOPNJBM&RVBUJPOTr6TFUIF5IFPSFNGPS #PVOETPO;FSPTr6TFUIF*OUFSNFEJBUF7BMVF5IFPSFN 5.6 Complex Zeros; Fundamental Theorem of Algebra 390 6TFUIF$POKVHBUF1BJST5IFPSFNr'JOEB1PMZOPNJBM'VODUJPOXJUI 4QFDJGJFE;FSPTr'JOEUIF$PNQMFY;FSPTPGB1PMZOPNJBM'VODUJPO 6 Chapter Review 396 Chapter Test 399 Cumulative Review 399 Chapter Projects 400 Exponential and Logarithmic Functions 6.1 Composite Functions 402 403 'PSNB$PNQPTJUF'VODUJPOr'JOEUIF%PNBJOPGB$PNQPTJUF'VODUJPO 6.2 One-to-One Functions; Inverse Functions 411 %FUFSNJOF8IFUIFSB'VODUJPO*T0OFUP0OFr%FUFSNJOFUIF*OWFSTFPGB 'VODUJPO%FGJOFECZB.BQPSB4FUPG0SEFSFE1BJSTr0CUBJOUIF(SBQIPG UIF*OWFSTF'VODUJPOGSPNUIF(SBQIPGUIF'VODUJPOr'JOEUIF*OWFSTFPGB Function Defined by an Equation 6.3 Exponential Functions &WBMVBUF&YQPOFOUJBM'VODUJPOTr(SBQI&YQPOFOUJBM'VODUJPOTr%FGJOF the Number er4PMWF&YQPOFOUJBM&RVBUJPOT 423 CONTENTS 6.4 Logarithmic Functions 440 Change Exponential Statements to Logarithmic Statements and -PHBSJUINJD4UBUFNFOUTUP&YQPOFOUJBM4UBUFNFOUTr&WBMVBUF-PHBSJUINJD &YQSFTTJPOTr%FUFSNJOFUIF%PNBJOPGB-PHBSJUINJD'VODUJPOr(SBQI -PHBSJUINJD'VODUJPOTr4PMWF-PHBSJUINJD&RVBUJPOT 6.5 Properties of Logarithms 452 8PSLXJUIUIF1SPQFSUJFTPG-PHBSJUINTr8SJUFB-PHBSJUINJD &YQSFTTJPOBTB4VNPS%JGGFSFODFPG-PHBSJUINTr8SJUFB-PHBSJUINJD &YQSFTTJPOBTB4JOHMF-PHBSJUINr&WBMVBUF-PHBSJUINT8IPTF#BTF*T Neither 10 Nor e 6.6 Logarithmic and Exponential Equations 461 4PMWF-PHBSJUINJD&RVBUJPOTr4PMWF&YQPOFOUJBM&RVBUJPOTr4PMWF Logarithmic and Exponential Equations Using a Graphing Utility 6.7 Financial Models 468 %FUFSNJOFUIF'VUVSF7BMVFPGB-VNQ4VNPG.POFZr$BMDVMBUF &GGFDUJWF3BUFTPG3FUVSOr%FUFSNJOFUIF1SFTFOU7BMVFPGB-VNQ 4VNPG.POFZr%FUFSNJOFUIF3BUFPG*OUFSFTUPSUIF5JNF3FRVJSFEUP Double a Lump Sum of Money 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models 478 Find Equations of Populations That Obey the Law of Uninhibited Growth r'JOE&RVBUJPOTPG1PQVMBUJPOT5IBU0CFZUIF-BXPG%FDBZr6TF/FXUPOT -BXPG$PPMJOHr6TF-PHJTUJD.PEFMT 6.9 Building Exponential, Logarithmic, and Logistic Models from Data 489 #VJMEBO&YQPOFOUJBM.PEFMGSPN%BUBr#VJMEB-PHBSJUINJD.PEFMGSPN %BUBr#VJMEB-PHJTUJD.PEFMGSPN%BUB 7 Chapter Review 497 Chapter Test 502 Cumulative Review 502 Chapter Projects 503 Trigonometric Functions 505 7.1 Angles and Their Measure 506 Convert between Decimal and Degree, Minute, Second Measures for "OHMFTr'JOEUIF-FOHUIPGBO"SDPGB$JSDMFr$POWFSUGSPN %FHSFFTUP3BEJBOTBOEGSPN3BEJBOTUP%FHSFFTr'JOEUIF"SFBPGB 4FDUPSPGB$JSDMFr'JOEUIF-JOFBS4QFFEPGBO0CKFDU5SBWFMJOHJO Circular Motion 7.2 Right Triangle Trigonometry 519 'JOEUIF7BMVFTPG5SJHPOPNFUSJD'VODUJPOTPG"DVUF"OHMFTr6TF 'VOEBNFOUBM*EFOUJUJFTr'JOEUIF7BMVFTPGUIF3FNBJOJOH5SJHPOPNFUSJD 'VODUJPOT (JWFOUIF7BMVFPG0OFPG5IFNr6TFUIF$PNQMFNFOUBSZ"OHMF Theorem 7.3 Computing the Values of Trigonometric Functions of Acute Angles p Find the Exact Values of the Trigonometric Functions of = 45°r'JOE 4 p p = 30° and = 60° the Exact Values of the Trigonometric Functions of 6 3 r6TFB$BMDVMBUPSUP"QQSPYJNBUFUIF7BMVFTPGUIF5SJHPOPNFUSJD 'VODUJPOTPG"DVUF"OHMFTr.PEFMBOE4PMWF"QQMJFE1SPCMFNT*OWPMWJOH Right Triangles 531 xi xii CONTENTS 7.4 Trigonometric Functions of Any Angle 543 'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD'VODUJPOTGPS"OZ"OHMFr6TF Coterminal Angles to Find the Exact Value of a Trigonometric Function r%FUFSNJOFUIF4JHOTPGUIF5SJHPOPNFUSJD'VODUJPOTPGBO"OHMFJOB(JWFO 2VBESBOUr'JOEUIF3FGFSFODF"OHMFPGBO"OHMFr6TFB3FGFSFODF"OHMF UP'JOEUIF&YBDU7BMVFPGB5SJHPOPNFUSJD'VODUJPOr'JOEUIF&YBDU7BMVFT of the Trigonometric Functions of an Angle, Given Information about the Functions 7.5 Unit Circle Approach; Properties of the Trigonometric Functions 553 Find the Exact Values of the Trigonometric Functions Using the Unit $JSDMFr,OPXUIF%PNBJOBOE3BOHFPGUIF5SJHPOPNFUSJD'VODUJPOTr6TF Periodic Properties to Find the Exact Values of the Trigonometric Functions r6TF&WFOm0EE1SPQFSUJFTUP'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD Functions 7.6 Graphs of the Sine and Cosine Functions 564 7.7 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions 579 7.8 Phase Shift; Sinusoidal Curve Fitting 587 Graph Functions of the Form y = A sin (vx) 6TJOH5SBOTGPSNBUJPOTr(SBQI Functions of the Form y = A cos (vx)6TJOH5SBOTGPSNBUJPOTr%FUFSNJOF UIF"NQMJUVEFBOE1FSJPEPG4JOVTPJEBM'VODUJPOTr(SBQI4JOVTPJEBM 'VODUJPOT6TJOH,FZ1PJOUTr'JOEBO&RVBUJPOGPSB4JOVTPJEBM(SBQI Graph Functions of the Form y = A tan (vx) + B and y = A cot (vx) + B r(SBQI'VODUJPOTPGUIF'PSNy = A csc (vx) + B and y = A sec (vx) + B Graph Sinusoidal Functions of the Form y = A sin (vx - f) + B r#VJME4JOVTPJEBM.PEFMTGSPN%BUB 8 Chapter Review 597 Chapter Test 603 Cumulative Review 603 Chapter Projects 604 Analytic Trigonometry 8.1 The Inverse Sine, Cosine, and Tangent Functions 606 607 'JOEUIF&YBDU7BMVFPGBO*OWFSTF4JOF'VODUJPOr'JOEBO"QQSPYJNBUF 7BMVFPGBO*OWFSTF4JOF'VODUJPOr6TF1SPQFSUJFTPG*OWFSTF'VODUJPOT UP'JOE&YBDU7BMVFTPG$FSUBJO$PNQPTJUF'VODUJPOTr'JOEUIF*OWFSTF 'VODUJPOPGB5SJHPOPNFUSJD'VODUJPOr4PMWF&RVBUJPOT*OWPMWJOH*OWFSTF Trigonometric Functions 8.2 The Inverse Trigonometric Functions (Continued) 620 Find the Exact Value of Expressions Involving the Inverse Sine, Cosine, and 5BOHFOU'VODUJPOTr%FGJOFUIF*OWFSTF4FDBOU $PTFDBOU BOE$PUBOHFOU 'VODUJPOTr6TFB$BMDVMBUPSUP&WBMVBUFsec -1x, csc -1x, and cot -1xr8SJUF a Trigonometric Expression as an Algebraic Expression 8.3 Trigonometric Equations 625 4PMWF&RVBUJPOT*OWPMWJOHB4JOHMF5SJHPOPNFUSJD'VODUJPOr4PMWF 5SJHPOPNFUSJD&RVBUJPOT6TJOHB$BMDVMBUPSr4PMWF5SJHPOPNFUSJD&RVBUJPOT 2VBESBUJDJO'PSNr4PMWF5SJHPOPNFUSJD&RVBUJPOT6TJOH'VOEBNFOUBM *EFOUJUJFTr4PMWF5SJHPOPNFUSJD&RVBUJPOT6TJOHB(SBQIJOH6UJMJUZ 8.4 Trigonometric Identities 6TF"MHFCSBUP4JNQMJGZ5SJHPOPNFUSJD&YQSFTTJPOTr&TUBCMJTI*EFOUJUJFT 635 CONTENTS 8.5 Sum and Difference Formulas xiii 643 6TF4VNBOE%JGGFSFODF'PSNVMBTUP'JOE&YBDU7BMVFTr6TF4VNBOE %JGGFSFODF'PSNVMBTUP&TUBCMJTI*EFOUJUJFTr6TF4VNBOE%JGGFSFODF 'PSNVMBT*OWPMWJOH*OWFSTF5SJHPOPNFUSJD'VODUJPOTr4PMWF5SJHPOPNFUSJD Equations Linear in Sine and Cosine 8.6 Double-angle and Half-angle Formulas 655 6TF%PVCMFBOHMF'PSNVMBTUP'JOE&YBDU7BMVFTr6TF%PVCMFBOHMF'PSNVMBT UP&TUBCMJTI*EFOUJUJFTr6TF)BMGBOHMF'PSNVMBTUP'JOE&YBDU7BMVFT 8.7 Product-to-Sum and Sum-to-Product Formulas 665 &YQSFTT1SPEVDUTBT4VNTr&YQSFTT4VNTBT1SPEVDUT 9 Chapter Review 669 Chapter Test 672 Cumulative Review 673 Chapter Projects 674 Applications of Trigonometric Functions 9.1 Applications Involving Right Triangles 675 676 4PMWF3JHIU5SJBOHMFTr4PMWF"QQMJFE1SPCMFNT 9.2 The Law of Sines 682 4PMWF4""PS"4"5SJBOHMFTr4PMWF44"5SJBOHMFTr4PMWF"QQMJFE1SPCMFNT 9.3 The Law of Cosines 692 4PMWF4"45SJBOHMFTr4PMWF4445SJBOHMFTr4PMWF"QQMJFE1SPCMFNT 9.4 Area of a Triangle 699 'JOEUIF"SFBPG4"45SJBOHMFTr'JOEUIF"SFBPG4445SJBOHMFT 9.5 Simple Harmonic Motion; Damped Motion; Combining Waves 705 #VJMEB.PEFMGPSBO0CKFDUJO4JNQMF)BSNPOJD.PUJPOr"OBMZ[F 4JNQMF)BSNPOJD.PUJPOr"OBMZ[FBO0CKFDUJO%BNQFE.PUJPO r(SBQIUIF4VNPG5XP'VODUJPOT 10 Chapter Review 714 Chapter Test 716 Cumulative Review 717 Chapter Projects 718 Polar Coordinates; Vectors 10.1 Polar Coordinates 719 720 1MPU1PJOUT6TJOH1PMBS$PPSEJOBUFTr$POWFSUGSPN1PMBS$PPSEJOBUFTUP 3FDUBOHVMBS$PPSEJOBUFTr$POWFSUGSPN3FDUBOHVMBS$PPSEJOBUFTUP1PMBS $PPSEJOBUFTr5SBOTGPSN&RVBUJPOTCFUXFFO1PMBSBOE3FDUBOHVMBS'PSNT 10.2 Polar Equations and Graphs 729 Identify and Graph Polar Equations by Converting to Rectangular &RVBUJPOTr5FTU1PMBS&RVBUJPOTGPS4ZNNFUSZr(SBQI1PMBS&RVBUJPOT by Plotting Points 10.3 The Complex Plane; De Moivre’s Theorem 1MPU1PJOUTJOUIF$PNQMFY1MBOFr$POWFSUB$PNQMFY/VNCFSCFUXFFO 3FDUBOHVMBS'PSNBOE1PMBS'PSNr'JOE1SPEVDUTBOE2VPUJFOUTPG $PNQMFY/VNCFSTJO1PMBS'PSNr6TF%F.PJWSFT5IFPSFNr'JOE Complex Roots 744 xiv CONTENTS 10.4 Vectors 752 (SBQI7FDUPSTr'JOEB1PTJUJPO7FDUPSr"EEBOE4VCUSBDU7FDUPST "MHFCSBJDBMMZr'JOEB4DBMBS.VMUJQMFBOEUIF.BHOJUVEFPGB7FDUPS r'JOEB6OJU7FDUPSr'JOEB7FDUPSGSPN*UT%JSFDUJPOBOE.BHOJUVEF r.PEFMXJUI7FDUPST 10.5 The Dot Product 766 'JOEUIF%PU1SPEVDUPG5XP7FDUPSTr'JOEUIF"OHMFCFUXFFO5XP7FDUPST r%FUFSNJOF8IFUIFS5XP7FDUPST"SF1BSBMMFMr%FUFSNJOF8IFUIFS5XP 7FDUPST"SF0SUIPHPOBMr%FDPNQPTFB7FDUPSJOUP5XP0SUIPHPOBM7FDUPST r$PNQVUF8PSL 11 Chapter Review 773 Chapter Test 776 Cumulative Review 776 Chapter Projects 777 Analytic Geometry 11.1 Conics 778 779 Know the Names of the Conics 11.2 The Parabola 780 "OBMZ[F1BSBCPMBTXJUI7FSUFYBUUIF0SJHJOr"OBMZ[F1BSBCPMBTXJUI Vertex at (h, k r4PMWF"QQMJFE1SPCMFNT*OWPMWJOH1BSBCPMBT 11.3 The Ellipse 789 "OBMZ[F&MMJQTFTXJUI$FOUFSBUUIF0SJHJOr"OBMZ[F&MMJQTFTXJUI Center at (h, k r4PMWF"QQMJFE1SPCMFNT*OWPMWJOH&MMJQTFT 11.4 The Hyperbola 799 "OBMZ[F)ZQFSCPMBTXJUI$FOUFSBUUIF0SJHJOr'JOEUIF"TZNQUPUFTPG B)ZQFSCPMBr"OBMZ[F)ZQFSCPMBTXJUI$FOUFSBU(h, k)r4PMWF"QQMJFE Problems Involving Hyperbolas 11.5 Rotation of Axes; General Form of a Conic 812 *EFOUJGZB$POJDr6TFB3PUBUJPOPG"YFTUP5SBOTGPSN&RVBUJPOT r"OBMZ[FBO&RVBUJPO6TJOHB3PUBUJPOPG"YFTr*EFOUJGZ$POJDTXJUIPVU a Rotation of Axes 11.6 Polar Equations of Conics 820 "OBMZ[FBOE(SBQI1PMBS&RVBUJPOTPG$POJDTr$POWFSUUIF1PMBS Equation of a Conic to a Rectangular Equation 11.7 Plane Curves and Parametric Equations 826 (SBQI1BSBNFUSJD&RVBUJPOTr'JOEB3FDUBOHVMBS&RVBUJPOGPSB $VSWF%FGJOFE1BSBNFUSJDBMMZr6TF5JNFBTB1BSBNFUFSJO1BSBNFUSJD &RVBUJPOTr'JOE1BSBNFUSJD&RVBUJPOTGPS$VSWFT%FGJOFECZ Rectangular Equations 12 Chapter Review 838 Chapter Test 841 Cumulative Review 841 Chapter Projects 842 Systems of Equations and Inequalities 12.1 Systems of Linear Equations: Substitution and Elimination 4PMWF4ZTUFNTPG&RVBUJPOTCZ4VCTUJUVUJPOr4PMWF4ZTUFNTPG&RVBUJPOT CZ&MJNJOBUJPOr*EFOUJGZ*ODPOTJTUFOU4ZTUFNTPG&RVBUJPOT$POUBJOJOH 843 844 CONTENTS xv 5XP7BSJBCMFTr&YQSFTTUIF4PMVUJPOPGB4ZTUFNPG%FQFOEFOU&RVBUJPOT $POUBJOJOH5XP7BSJBCMFTr4PMWF4ZTUFNTPG5ISFF&RVBUJPOT$POUBJOJOH 5ISFF7BSJBCMFTr*EFOUJGZ*ODPOTJTUFOU4ZTUFNTPG&RVBUJPOT$POUBJOJOH 5ISFF7BSJBCMFTr&YQSFTTUIF4PMVUJPOPGB4ZTUFNPG%FQFOEFOU&RVBUJPOT Containing Three Variables 12.2 Systems of Linear Equations: Matrices 859 8SJUFUIF"VHNFOUFE.BUSJYPGB4ZTUFNPG-JOFBS&RVBUJPOTr8SJUFUIF 4ZTUFNPG&RVBUJPOTGSPNUIF"VHNFOUFE.BUSJYr1FSGPSN3PX0QFSBUJPOT POB.BUSJYr4PMWFB4ZTUFNPG-JOFBS&RVBUJPOT6TJOH.BUSJDFT 12.3 Systems of Linear Equations: Determinants 874 &WBMVBUFCZ%FUFSNJOBOUTr6TF$SBNFST3VMFUP4PMWFB4ZTUFNPG 5XP&RVBUJPOT$POUBJOJOH5XP7BSJBCMFTr&WBMVBUFCZ%FUFSNJOBOUT r6TF$SBNFST3VMFUP4PMWFB4ZTUFNPG5ISFF&RVBUJPOT$POUBJOJOH5ISFF 7BSJBCMFTr,OPX1SPQFSUJFTPG%FUFSNJOBOUT 12.4 Matrix Algebra 884 'JOEUIF4VNBOE%JGGFSFODFPG5XP.BUSJDFTr'JOE4DBMBS.VMUJQMFTPGB .BUSJYr'JOEUIF1SPEVDUPG5XP.BUSJDFTr'JOEUIF*OWFSTFPGB.BUSJY Solve a System of Linear Equations Using an Inverse Matrix 12.5 Partial Fraction Decomposition 901 P Decompose , Where Q Has Only Nonrepeated Linear Factors Q P P r%FDPNQPTF , Where Q)BT3FQFBUFE-JOFBS'BDUPSTr%FDPNQPTF , Q Q P Where Q)BTB/POSFQFBUFE*SSFEVDJCMF2VBESBUJD'BDUPSr%FDPNQPTF , Q Where Q Has a Repeated Irreducible Quadratic Factor 12.6 Systems of Nonlinear Equations 909 4PMWFB4ZTUFNPG/POMJOFBS&RVBUJPOT6TJOH4VCTUJUVUJPOr4PMWFB System of Nonlinear Equations Using Elimination 12.7 Systems of Inequalities 918 (SBQIBO*OFRVBMJUZr(SBQIB4ZTUFNPG*OFRVBMJUJFT 12.8 Linear Programming 925 4FU6QB-JOFBS1SPHSBNNJOH1SPCMFNr4PMWFB-JOFBS1SPHSBNNJOH Problem 13 Chapter Review 932 Chapter Test 936 Cumulative Review 937 Chapter Projects 937 Sequences; Induction; the Binomial Theorem 13.1 Sequences 939 940 8SJUFUIF'JSTU4FWFSBM5FSNTPGB4FRVFODFr8SJUFUIF5FSNTPGB 4FRVFODF%FGJOFECZB3FDVSTJWF'PSNVMBr6TF4VNNBUJPO/PUBUJPO r'JOEUIF4VNPGB4FRVFODF 13.2 Arithmetic Sequences 950 %FUFSNJOF8IFUIFSB4FRVFODF*T"SJUINFUJDr'JOEB'PSNVMBGPSBO "SJUINFUJD4FRVFODFr'JOEUIF4VNPGBO"SJUINFUJD4FRVFODF 13.3 Geometric Sequences; Geometric Series %FUFSNJOF8IFUIFSB4FRVFODF*T(FPNFUSJDr'JOEB'PSNVMBGPSB (FPNFUSJD4FRVFODFr'JOEUIF4VNPGB(FPNFUSJD4FRVFODF r%FUFSNJOF8IFUIFSB(FPNFUSJD4FSJFT$POWFSHFTPS%JWFSHFTr4PMWF Annuity Problems 956 xvi CONTENTS 13.4 Mathematical Induction 967 Prove Statements Using Mathematical Induction 13.5 The Binomial Theorem 971 n Evaluate a b r6TFUIF#JOPNJBM5IFPSFN j 14 Chapter Review 977 Chapter Test 979 Cumulative Review 980 Chapter Projects 981 Counting and Probability 14.1 Counting 982 983 'JOE"MMUIF4VCTFUTPGB4FUr$PVOUUIF/VNCFSPG&MFNFOUTJOB4FU r4PMWF$PVOUJOH1SPCMFNT6TJOHUIF.VMUJQMJDBUJPO1SJODJQMF 14.2 Permutations and Combinations 988 Solve Counting Problems Using Permutations Involving n Distinct 0CKFDUTr4PMWF$PVOUJOH1SPCMFNT6TJOH$PNCJOBUJPOTr4PMWF$PVOUJOH Problems Using Permutations Involving n Nondistinct Objects 14.3 Probability 997 $POTUSVDU1SPCBCJMJUZ.PEFMTr$PNQVUF1SPCBCJMJUJFTPG&RVBMMZ-JLFMZ 0VUDPNFTr'JOE1SPCBCJMJUJFTPGUIF6OJPOPG5XP&WFOUTr6TFUIF Complement Rule to Find Probabilities Appendix Chapter Review 1007 Chapter Test 1009 Cumulative Review 1010 Chapter Projects 1010 Graphing Utilities A1 A.1 The Viewing Rectangle A1 A.2 Using a Graphing Utility to Graph Equations A3 A.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry A5 A.4 Using a Graphing Utility to Solve Equations A6 A.5 Square Screens A8 A.6 Using a Graphing Utility to Graph Inequalities A9 A.7 Using a Graphing Utility to Solve Systems of Linear Equations A9 A.8 Using a Graphing Utility to Graph a Polar Equation A11 A.9 Using a Graphing Utility to Graph Parametric Equations A11 Answers AN1 Credits Index C1 I1 For the family Katy (Murphy) and Pat Shannon, Patrick, Ryan Mike and Yola Michael, Kevin, Marissa Dan and Sheila Maeve, Sean, Nolan Colleen (O’Hara) and Bill Kaleigh, Billy, Timmy Three Distinct Series Students have different goals, learning styles, and levels of preparation. Instructors have different teaching philosophies, styles, and techniques. Rather than write one series to fit all, the Sullivans have written three distinct series. All share the same goal—to develop a high level of mathematical understanding and an appreciation for the way mathematics can describe the world around us. The manner of reaching that goal, however, differs from series to series. Contemporary Series, Tenth Edition The Contemporary Series is the most traditional in approach yet modern in its treatment of precalculus mathematics. Graphing utility coverage is optional and can be included or excluded at the discretion of the instructor: College Algebra, Algebra & Trigonometry, Trigonometry: A Unit Circle Approach, Precalculus. Enhanced with Graphing Utilities Series, Sixth Edition This series provides a thorough integration of graphing utilities into topics, allowing students to explore mathematical concepts and encounter ideas usually studied in later courses. Using technology, the approach to solving certain problems differs from the Contemporary Series, while the emphasis on understanding concepts and building strong skills does not: College Algebra, Algebra & Trigonometry, Precalculus. Concepts through Functions Series, Third Edition This series differs from the others, utilizing a functions approach that serves as the organizing principle tying concepts together. Functions are introduced early in various formats. This approach supports the Rule of Four, which states that functions are represented symbolically, numerically, graphically, and verbally. Each chapter introduces a new type of function and then develops all concepts pertaining to that particular function. The solutions of equations and inequalities, instead of being developed as stand-alone topics, are developed in the context of the underlying functions. Graphing utility coverage is optional and can be included or excluded at the discretion of the instructor: College Algebra; Precalculus, with a Unit Circle Approach to Trigonometry; Precalculus, with a Right Triangle Approach to Trigonometry. xviii The Contemporary Series College Algebra, Tenth Edition This text provides a contemporary approach to college algebra, with three chapters of review material preceding the chapters on functions. Graphing calculator usage is provided, but is optional. After completing this book, a student will be adequately prepared for trigonometry, finite mathematics, and business calculus. Algebra & Trigonometry, Tenth Edition This text contains all the material in College Algebra, but also develops the trigonometric functions using a right triangle approach and showing how it relates to the unit circle approach. Graphing techniques are emphasized, including a thorough discussion of polar coordinates, parametric equations, and conics using polar coordinates. Graphing calculator usage is provided, but is optional. After completing this book, a student will be adequately prepared for finite mathematics, business calculus, and engineering calculus. Precalculus, Tenth Edition This text contains one review chapter before covering the traditional precalculus topic of functions and their graphs, polynomial and rational functions, and exponential and logarithmic functions. The trigonometric functions are introduced using a unit circle approach and showing how it relates to the right triangle approach. Graphing techniques are emphasized, including a thorough discussion of polar coordinates, parametric equations, and conics using polar coordinates. Graphing calculator usage is provided, but is optional. The final chapter provides an introduction to calculus, with a discussion of the limit, the derivative, and the integral of a function. After completing this book, a student will be adequately prepared for finite mathematics, business calculus, and engineering calculus. Trigonometry: a Unit Circle Approach, Tenth Edition This text, designed for stand-alone courses in trigonometry, develops the trigonometric functions using a unit circle approach and showing how it relates to the right triangle approach. Graphing techniques are emphasized, including a thorough discussion of polar coordinates, parametric equations, and conics using polar coordinates. Graphing calculator usage is provided, but is optional. After completing this book, a student will be adequately prepared for finite mathematics, business calculus, and engineering calculus. xix Preface to the Instructor A s a professor of mathematics at an urban public university for 35 years, I understand the varied needs of algebra and trigonometry students. Students range from being underprepared, with little mathematical background and a fear of mathematics, to being highly prepared and motivated. For some, this is their final course in mathematics. For others, it is preparation for future mathematics courses. I have written this text with both groups in mind. A tremendous benefit of authoring a successful series is the broad-based feedback I receive from teachers and students who have used previous editions. I am sincerely grateful for their support. Virtually every change to this edition is the result of their thoughtful comments and suggestions. I hope that I have been able to take their ideas and, building upon a successful foundation of the ninth edition, make this series an even better learning and teaching tool for students and teachers. Features in the Tenth Edition A descriptive list of the many special features of Algebra & Trigonometry can be found on the endpapers in the front of this text. This list places the features in their proper context, as building blocks of an overall learning system that has been carefully crafted over the years to help students get the most out of the time they put into studying. Please take the time to review this and to discuss it with your students at the beginning of your course. My experience has been that when students utilize these features, they are more successful in the course. New to the Tenth Edition r Retain Your Knowledge This new category of problems in the exercise set are based on the article “To Retain New Learning, Do the Math” published in the Edurati Review. In this article, Kevin Washburn suggests that “the more students are required to recall new content or skills, the better their memory will be.” It is frustrating when students cannot recall skills learned earlier in the course. To alleviate this recall problem, we have created “Retain Your Knowledge” problems. These are problems considered to be “final exam material” that students can use to maintain their skills. All the answers to these problems appear in the back of the text, and all are programmed in MyMathLab. r Guided Lecture Notes Ideal for online, emporium/ redesign courses, inverted classrooms, or traditional lecture classrooms. These lecture notes help students take thorough, organized, and understandable notes as they watch the Author in Action videos. They ask students to complete definitions, procedures, and examples based on the content of the videos and text. In addition, experience suggests that students learn by doing and understanding the why/how of the concept or xx r r r r property.Therefore, many sections will have an exploration activity to motivate student learning. These explorations introduce the topic and/or connect it to either a real-world application or a previous section. For example, when the vertical-line test is discussed in Section 3.2, after the theorem statement, the notes ask the students to explain why the vertical-line test works by using the definition of a function. This challenge helps students process the information at a higher level of understanding. Illustrations Many of the figures now have captions to help connect the illustrations to the explanations in the body of the text. TI Screen Shots In this edition we have replaced all the screen shots from the ninth edition with screen shots using TI-84Plus C. These updated screen shots help students visualize concepts clearly and help make stronger connections between equations, data, and graphs in full color. Chapter Projects, which apply the concepts of each chapter to a real-world situation, have been enhanced to give students an up-to-the-minute experience. Many projects are new and Internet-based, requiring the student to research information online in order to solve problems. Exercise Sets All the exercises in the text have been reviewed and analyzed for this edition, some have been removed, and new ones have been added. All time-sensitive problems have been updated to the most recent information available. The problem sets remain classified according to purpose. The ‘Are You Prepared?’ problems have been improved to better serve their purpose as a just-in-time review of concepts that the student will need to apply in the upcoming section. The Concepts and Vocabulary problems have been expanded and now include multiple-choice exercises. Together with the fill-in-the-blank and True/False problems, these exercises have been written to serve as reading quizzes. Skill Building problems develop the student’s computational skills with a large selection of exercises that are directly related to the objectives of the section. Mixed Practice problems offer a comprehensive assessment of skills that relate to more than one objective. Often these require skills learned earlier in the course. Applications and Extensions problems have been updated. Further, many new application-type exercises have been added, especially ones involving information and data drawn from sources the student will recognize, to improve relevance and timeliness. The Explaining Concepts: Discussion and Writing exercises have been improved and expanded to provide more opportunity for classroom discussion and group projects. PREFACE xxi New to this edition, Retain Your Knowledge exercises consist of a collection of four problems in each exercise set that are based on material learned earlier in the course. They serve to keep information that has already been learned “fresh” in the mind of the student. Answers to all these problems appear in the Student Edition. The Review Exercises in the Chapter Review have been streamlined, but they remain tied to the clearly expressed objectives of the chapter. Answers to all these problems appear in the Student Edition. r Annotated Instructor’s Edition As a guide, the author’s suggestions for homework assignments are indicated by a blue underscore below the problem number. These problems are assignable in the MyMathLab as part of a “Ready-to-Go” course. Chapter R Review This chapter consists of review material. It may be used as the first part of the course or later as a just-in-time review when the content is required. Specific references to this chapter occur throughout the text to assist in the review process. Content Changes in the Tenth Edition Chapter 3 Functions and Their Graphs Perhaps the most important chapter. Section 3.6 is optional. r Section 3.1 The objective Find the Difference Quotient of a Function has been added. r Section 5.1 The subsection Behavior of the Graph of a 1PMZOPNJBM'VODUJPO/FBSB;FSPIBTCFFOSFNPWFE r Section 5.3 A subsection has been added that discusses the role of multiplicity of the zeros of the denominator of a rational function as it relates to the graph near a vertical asymptote. r Section 5.5 The objective Use Descartes’ Rule of Signs has been included. r Section 5.5 5IF UIFPSFN #PVOET PO UIF ;FSPT PG B Polynomial Function is now based on the traditional method of using synthetic division. Chapter 4 Linear and Quadratic Functions Topic selection depends on your syllabus. Sections 4.2 and 4.4 may be omitted without loss of continuity. Using the Tenth Edition Effectively with Your Syllabus To meet the varied needs of diverse syllabi, this text contains more content than is likely to be covered in an Algebra & Trigonometry course.As the chart illustrates, this text has been organized with flexibility of use in mind.Within a given chapter, certain sections are optional (see the details that follow the figure below) and can be omitted without loss of continuity. 1 R 4 5 6 11.111.4 12 14 7 13 8 9 10.110.3 10.410.5 Chapter 2 Graphs This chapter lays the foundation for functions. Section 2.5 is optional. Chapter 5 Polynomial and Rational Functions Topic selection depends on your syllabus. Chapter 6 Exponential and Logarithmic Functions 4FDUJPOT m GPMMPX JO TFRVFODF 4FDUJPOT and 6.9 are optional. Chapter 7 Trigonometric Functions Section 7.8 may be omitted in a brief course. Chapter 8 Analytic Trigonometry Sections 8.2, 8.6, and 8.8 may be omitted in a brief course. Chapter 9 Applications of Trigonometric Functions Sections 9.4 and 9.5 may be omitted in a brief course. Chapter 10 Polar Coordinates; Vectors 4FDUJPOTmBOE4FDUJPOTmBSFJOEFQFOEFOU and may be covered separately. Chapter 11 Analytic Geometry 4FDUJPOT m GPMMPX JO TFRVFODF 4FDUJPOT and 11.7 are independent of each other, but each requires 4FDUJPOTm Chapter 12 Systems of Equations and Inequalities 4FDUJPOTmNBZCFDPWFSFEJOBOZPSEFS CVUFBDI requires Section 12.1. Section 12.8 requires Section 12.7. 2 3 Chapter 1 Equations and Inequalities Primarily a review of Intermediate Algebra topics, this material is a prerequisite for later topics. The coverage of complex numbers and quadratic equations with a negative discriminant is optional and may be postponed or skipped entirely without loss of continuity. 11.511.7 Chapter 13 Sequences; Induction; The Binomial Theorem 5IFSF BSF UISFF JOEFQFOEFOU QBSUT 4FDUJPOT m Section 13.4; and Section 13.5. Chapter 14 Counting and Probability The sections follow in sequence. xxii PREFACE Acknowledgments Textbooks are written by authors, but evolve from an idea to final form through the efforts of many people. It was Don Dellen who first suggested this text and series to me. Don is remembered for his extensive contributions to publishing and mathematics. Thanks are due to the following people for their assistance and encouragement to the preparation of this edition: r From Pearson Education: Anne Kelly for her substantial contributions, ideas, and enthusiasm; Dawn Murrin, for her unmatched talent at getting the details right; Joseph Colella for always getting the reviews and pages to me on time; Peggy McMahon for directing the always difficult production process; Rose Kernan for handling James Africh, College of DuPage Steve Agronsky, Cal Poly State University Gererdo Aladro, Florida International University Grant Alexander, Joliet Junior College Dave Anderson, South Suburban College Richard Andrews, Florida A&M University Joby Milo Anthony, University of Central Florida James E. Arnold, University of Wisconsin-Milwaukee Adel Arshaghi, Center for Educational Merit Carolyn Autray, University of West Georgia Agnes Azzolino, Middlesex County College Wilson P. Banks, Illinois State University Sudeshna Basu, Howard University Dale R. Bedgood, East Texas State University Beth Beno, South Suburban College Carolyn Bernath, Tallahassee Community College Rebecca Berthiaume, Edison State College William H. Beyer, University of Akron Annette Blackwelder, Florida State University Richelle Blair, Lakeland Community College Kevin Bodden, Lewis and Clark College Jeffrey Boerner, University of Wisconsin-Stout Barry Booten, Florida Atlantic University Larry Bouldin, Roane State Community College Bob Bradshaw, Ohlone College Trudy Bratten, Grossmont College Tim Bremer, Broome Community College Tim Britt, Jackson State Community College Michael Brook, University of Delaware Joanne Brunner, Joliet Junior College Warren Burch, Brevard Community College Mary Butler, Lincoln Public Schools Melanie Butler, West Virginia University Jim Butterbach, Joliet Junior College William J. Cable, University of Wisconsin-Stevens Point Lois Calamia, Brookdale Community College Jim Campbell, Lincoln Public Schools Roger Carlsen, Moraine Valley Community College Elena Catoiu, Joliet Junior College Mathews Chakkanakuzhi, Palomar College Tim Chappell, Penn Valley Community College John Collado, South Suburban College Alicia Collins, Mesa Community College Nelson Collins, Joliet Junior College Rebecca Connell, Troy University Jim Cooper, Joliet Junior College Denise Corbett, East Carolina University liaison between the compositor and author; Peggy Lucas for her genuine interest in marketing this text; Chris Hoag for her continued support and genuine interest; Greg Tobin for his leadership and commitment to excellence; and the Pearson Math and Science Sales team, for their continued confidence and personal support of our texts. r Accuracy checkers: C. Brad Davis, who read the entire manuscript and accuracy checked answers. His attention to detail is amazing; Timothy Britt, for creating the Solutions Manuals and accuracy checking answers. Finally, I offer my grateful thanks to the dedicated users and reviewers of my texts, whose collective insights form the backbone of each textbook revision. Carlos C. Corona, San Antonio College Theodore C. Coskey, South Seattle Community College Rebecca Connell, Troy University Donna Costello, Plano Senior High School Paul Crittenden, University of Nebraska at Lincoln John Davenport, East Texas State University Faye Dang, Joliet Junior College Antonio David, Del Mar College Stephanie Deacon, Liberty University Duane E. Deal, Ball State University Jerry DeGroot, Purdue North Central Timothy Deis, University of WisconsinPlatteville Joanna DelMonaco, Middlesex Community College Vivian Dennis, Eastfield College Deborah Dillon, R. L. Turner High School Guesna Dohrman, Tallahassee Community College Cheryl Doolittle, Iowa State University Karen R. Dougan, University of Florida Jerrett Dumouchel, Florida Community College at Jacksonville Louise Dyson, Clark College Paul D. East, Lexington Community College Don Edmondson, University of Texas-Austin Erica Egizio, Joliet Junior College Jason Eltrevoog, Joliet Junior College Christopher Ennis, University of Minnesota Kathy Eppler, Salt Lake Community College Ralph Esparza, Jr., Richland College Garret J. Etgen, University of Houston Scott Fallstrom, Shoreline Community College Pete Falzone, Pensacola Junior College Arash Farahmand, Skyline College W.A. Ferguson, University of Illinois-Urbana/ Champaign Iris B. Fetta, Clemson University Mason Flake, student at Edison Community College Timothy W. Flood, Pittsburg State University Robert Frank, Westmoreland County Community College Merle Friel, Humboldt State University Richard A. Fritz, Moraine Valley Community College Dewey Furness, Ricks College Mary Jule Gabiou, North Idaho College Randy Gallaher, Lewis and Clark College Tina Garn, University of Arizona Dawit Getachew, Chicago State University Wayne Gibson, Rancho Santiago College Loran W. Gierhart, University of Texas at San Antonio and Palo Alto College Robert Gill, University of Minnesota Duluth Nina Girard, University of Pittsburgh at Johnstown Sudhir Kumar Goel, Valdosta State University Adrienne Goldstein, Miami Dade College, Kendall Campus Joan Goliday, Sante Fe Community College Lourdes Gonzalez, Miami Dade College, Kendall Campus Frederic Gooding, Goucher College Donald Goral, Northern Virginia Community College Sue Graupner, Lincoln Public Schools Mary Beth Grayson, Liberty University Jennifer L. Grimsley, University of Charleston Ken Gurganus, University of North Carolina James E. Hall, University of Wisconsin-Madison Judy Hall, West Virginia University Edward R. Hancock, DeVry Institute of Technology Julia Hassett, DeVry Institute, Dupage Christopher Hay-Jahans, University of South Dakota Michah Heibel, Lincoln Public Schools LaRae Helliwell, San Jose City College Celeste Hernandez, Richland College Gloria P. Hernandez, Louisiana State University at Eunice Brother Herron, Brother Rice High School Robert Hoburg, Western Connecticut State University Lynda Hollingsworth, Northwest Missouri State University Deltrye Holt, Augusta State University Charla Holzbog, Denison High School Lee Hruby, Naperville North High School Miles Hubbard, St. Cloud State University Kim Hughes, California State College-San Bernardino Stanislav, Jabuka, University of Nevada, Reno Ron Jamison, Brigham Young University Richard A. Jensen, Manatee Community College Glenn Johnson, Middlesex Community College Sandra G. Johnson, St. Cloud State University Tuesday Johnson, New Mexico State University Susitha Karunaratne, Purdue University North Central Moana H. Karsteter, Tallahassee Community College Donna Katula, Joliet Junior College PREFACE Arthur Kaufman, College of Staten Island Thomas Kearns, North Kentucky University +BDL,FBUJOH .BTTBTPJU$PNNVOJUZ$PMMFHF Shelia Kellenbarger, Lincoln Public Schools Rachael Kenney, North Carolina State University +PIO#,MBTTFO /PSUI*EBIP$PMMFHF Debra Kopcso, Louisiana State University Lynne Kowski, Raritan Valley Community College Yelena Kravchuk, University of Alabama at Birmingham Ray S. Kuan, Skyline College Keith Kuchar, Manatee Community College Tor Kwembe, Chicago State University -JOEB+,ZMF 5BSSBOU$PVOUSZ+S$PMMFHF H.E. Lacey, Texas A & M University Harriet Lamm, Coastal Bend College +BNFT-BQQ 'PSU-FXJT$PMMFHF Matt Larson, Lincoln Public Schools Christopher Lattin, Oakton Community College +VMJB-FEFU -PVTJBOB4UBUF6OJWFSTJUZ Adele LeGere, Oakton Community College Kevin Leith, University of Houston +P"OO-FXJO &EJTPO$PMMFHF +FGG-FXJT +PIOTPO$PVOUZ$PNNVOJUZ$PMMFHF +BOJDF$-ZPO 5BMMBIBTTFF$PNNVOJUZ$PMMFHF +FBO.D"SUIVS +PMJFU+VOJPS$PMMFHF Virginia McCarthy, Iowa State University Karla McCavit, Albion College Michael McClendon, University of Central Oklahoma Tom McCollow, DeVry Institute of Technology Marilyn McCollum, North Carolina State University +JMM.D(PXBO )PXBSE6OJWFSTJUZ Will McGowant, Howard University "OHFMB.D/VMUZ +PMJFU+VOJPS$PMMFHF Laurence Maher, North Texas State University +BZ".BMNTUSPN 0LMBIPNB$JUZ$PNNVOJUZ College Rebecca Mann, Apollo High School Lynn Marecek, Santa Ana College Sherry Martina, Naperville North High School Alec Matheson, Lamar University Nancy Matthews, University of Oklahoma +BNFT.BYXFMM 0LMBIPNB4UBUF University-Stillwater Marsha May, Midwestern State University +BNFT.D-BVHIMJO 8FTU$IFTUFS6OJWFSTJUZ +VEZ.FDLMFZ +PMJFU+VOJPS$PMMFHF David Meel, Bowling Green State University Carolyn Meitler, Concordia University Samia Metwali, Erie Community College 3JDI.FZFST +PMJFU+VOJPS$PMMFHF Eldon Miller, University of Mississippi +BNFT.JMMFS 8FTU7JSHJOJB6OJWFSTJUZ Michael Miller, Iowa State University Kathleen Miranda, SUNY at Old Westbury Chris Mirbaha, The Community College of Baltimore County Val Mohanakumar, Hillsborough Community College Thomas Monaghan, Naperville North High School Miguel Montanez, Miami Dade College, Wolfson Campus Maria Montoya, Our Lady of the Lake University Susan Moosai, Florida Atlantic University Craig Morse, Naperville North High School Samad Mortabit, Metropolitan State University Pat Mower, Washburn University Tammy Muhs, University of Central Florida A. Muhundan, Manatee Community College +BOF.VSQIZ .JEEMFTFY$PNNVOJUZ$PMMFHF Richard Nadel, Florida International University Gabriel Nagy, Kansas State University Bill Naegele, South Suburban College Karla Neal, Lousiana State University Lawrence E. Newman, Holyoke Community College Dwight Newsome, Pasco-Hernando Community College Denise Nunley, Maricopa Community Colleges +BNFT/ZNBOO 6OJWFSTJUZPG5FYBT&M1BTP Mark Omodt, Anoka-Ramsey Community College Seth F. Oppenheimer, Mississippi State University Leticia Oropesa, University of Miami -JOEB1BEJMMB +PMJFU+VOJPS$PMMFHF Sanja Pantic, University of Illinois at Chicago &+BNFT1FBLF *PXB4UBUF6OJWFSTJUZ Kelly Pearson, Murray State University Dashamir Petrela, Florida Atlantic University Philip Pina, Florida Atlantic University Charlotte Pisors, Baylor University Michael Prophet, University of Northern Iowa Laura Pyzdrowski, West Virginia University Carrie Quesnell, Weber State University Neal C. Raber, University of Akron 5IPNBT3BEJO 4BO+PBRVJO%FMUB$PMMFHF Aibeng Serene Radulovic, Florida Atlantic University Ken A. Rager, Metropolitan State College Kenneth D. Reeves, San Antonio College Elsi Reinhardt, Truckee Meadows Community College +PTF3FNFTBS .JBNJ%BEF$PMMFHF 8PMGTPO Campus +BOF3JOHXBME *PXB4UBUF6OJWFSTJUZ Douglas F. Robertson, University of Minnesota, MPLS Stephen Rodi, Austin Community College William Rogge, Lincoln Northeast High School Howard L. Rolf, Baylor University Mike Rosenthal, Florida International University Phoebe Rouse, Lousiana State University Edward Rozema, University of Tennessee at Chattanooga Dennis C. Runde, Manatee Community College Alan Saleski, Loyola University of Chicago 4VTBO4BOENFZFS +BNFTUPXO$PNNVOJUZ College Brenda Santistevan, Salt Lake Community College Linda Schmidt, Greenville Technical College Ingrid Scott, Montgomery College A.K. Shamma, University of West Florida ;BDIFSZ4IBSPO 6OJWFSTJUZPG5FYBTBU4BO Antonio Martin Sherry, Lower Columbia College Carmen Shershin, Florida International University 5BUSBOB4IVCJO 4BO+PTF4UBUF6OJWFSTJUZ Anita Sikes, Delgado Community College Timothy Sipka, Alma College xxiii Charlotte Smedberg, University of Tampa Lori Smellegar, Manatee Community College Gayle Smith, Loyola Blakefield Cindy Soderstrom, Salt Lake Community College Leslie Soltis, Mercyhurst College +PIO4QFMMNBO 4PVUIXFTU5FYBT4UBUF University Karen Spike, University of North Carolina Rajalakshmi Sriram, Okaloosa-Walton Community College Katrina Staley, North Carolina Agricultural and Technical State University Becky Stamper, Western Kentucky University +VEZ4UBWFS 'MPSJEB$PNNVOJUZ College-South Robin Steinberg, Pima Community College Neil Stephens, Hinsdale South High School Sonya Stephens, Florida A&M Univeristy 1BUSJDL4UFWFOT +PMJFU+VOJPS$PMMFHF +PIO4VNOFS 6OJWFSTJUZPG5BNQB Matthew TenHuisen, University of North Carolina, Wilmington Christopher Terry, Augusta State University Diane Tesar, South Suburban College Tommy Thompson, Brookhaven College Martha K. Tietze, Shawnee Mission Northwest High School 3JDIBSE+5POESB *PXB4UBUF6OJWFSTJUZ Florentina Tone, University of West Florida Suzanne Topp, Salt Lake Community College Marilyn Toscano, University of Wisconsin, Superior Marvel Townsend, University of Florida +JN5SVEOPXTLJ $BSSPMM$PMMFHF 3PCFSU5VTLFZ +PMJFU+VOJPS$PMMFHF Mihaela Vajiac, Chapman University-Orange +VMJB7BSCBMPX 5IPNBT/FMTPO$PNNVOJUZ College-Leesville Richard G. Vinson, University of South Alabama +PSHF7JPMB1SJPMJ 'MPSJEB"UMBOUJD6OJWFSTJUZ Mary Voxman, University of Idaho +FOOJGFS8BMTI %BZUPOB#FBDI$PNNVOJUZ College Donna Wandke, Naperville North High School Timothy L.Warkentin, Cloud County Community College .FMJTTB+8BUUT 7JSHJOJB4UBUF6OJWFSTJUZ Hayat Weiss, Middlesex Community College Kathryn Wetzel, Amarillo College Darlene Whitkenack, Northern Illinois University Suzanne Williams, Central Piedmont Community College Larissa Williamson, University of Florida Christine Wilson, West Virginia University Brad Wind, Florida International University Anna Wiodarczyk, Florida International University Mary Wolyniak, Broome Community College Canton Woods, Auburn University Tamara S. Worner, Wayne State College Terri Wright, New Hampshire Community Technical College, Manchester "MFUIFJB;BNCFTJ 6OJWFSTJUZPG8FTU'MPSJEB (FPSHF;B[J $IJDBHP4UBUF6OJWFSTJUZ 4UFWF;VSP +PMJFU+VOJPS$PMMFHF Chicago State University Resources for Success Online Course (access code required) MyMathLab delivers proven results in helping individual students succeed. It provides engaging experiences that personalize, stimulate, and measure learning for each student. And it comes from an experienced partner with educational expertise and an eye on the future. MyMathLab helps prepare students and gets them thinking more conceptually and visually through the following features: Adaptive Study Plan The Study Plan makes studying more efficient and effective for every student. Performance and activity are assessed continually in real time. The data and analytics are used to provide personalized content– reinforcing concepts that target each student’s strengths and weaknesses. Getting Ready Students refresh prerequisite topics through assignable skill review quizzes and personalized homework integrated in MyMathLab. Video Assessment Video assessment is tied to key Author in Action videos to check students’ conceptual understanding of important math concepts. Enhanced Graphing Functionality New functionality within the graphing utility allows graphing of 3-point quadratic functions, 4-point cubic graphs, and transformations in exercises. Skills for Success Modules are integrated within the MyMathLab course to help students succeed in collegiate courses and prepare for future professions. Retain Your Knowledge These new exercises support ongoing review at the course level and help students maintain essential skills. xxiv Instructor Resources Student Resources Additional resources can be downloaded from www.pearsonhighered.com or hardcopy resources can be ordered from your sales representative. Additional resources to enhance student success: Ready to Go MyMathLab® Course Now it is even easier to get started with MyMathLab. The Ready to Go MyMathLab course option includes author-chosen preassigned homework, integrated review, and more. TestGen® TestGen® (www.pearsoned.com/testgen) enables instructors to build, edit, print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text. PowerPoint® Lecture Slides Lecture Video Author in Action videos are actual classroom lectures with fully worked out examples presented by Michael Sullivan, III. All video is assignable within MyMathlab. Chapter Test Prep Videos Students can watch instructors work through step-by-step solutions to all chapter test exercises from the text. These are available in MyMathlab and on YouTube. Fully editable slides correlated with the text. Student Solutions Manual Annotated Instructor’s Edition Provides detailed worked-out solutions to oddnumbered exercises. Shorter answers are on the page beside the exercises. Longer answers are in the back of the text. Guided Lecture Notes Includes additional examples and helpful teaching tips, by section. These lecture notes assist students in taking thorough, organized, and understandable notes while watching Author in Action videos. Students actively participate in learning the how/why of important concepts through explorations and activities. The Guided Lecture Notes are available as PDF’s and customizable Word files in MyMathLab. They can also be packaged with the text and the MyMathLab access code. Online Chapter Projects Algebra Review Additional projects that give students an opportunity to apply what they learned in the chapter. Four chapters of Intermediate Algebra review. Perfect for a slower-paced course or for individual review. Instructor Solutions Manual Includes fully worked solutions to all exercises in the text. Mini Lecture Notes xxv Applications Index Acoustics Area. See also Geometry amplifying sound, 500 loudness of sound, 451 loudspeaker, 712 tuning fork, 712, 713 whispering galleries, 795–796 of Bermuda Triangle, 703 under a curve, 619 of isosceles triangle, 664 of portion of rectangle outside of circle, 518 of sector of circle, 513, 516 of segment of circle, 715 of sidewalk, 531 for tethered dog to roam, 518 of windshield wiper sweep, 516 Aerodynamics modeling aircraft motion, 777 Aeronautics Challenger disaster, 488 Agriculture farm management, 931 farm workers in U.S., 487 field enclosure, 916 grazing area for cow, 704 milk production, 494 minimizing cost, 931 removing stump, 765 watering a field, 102 Air travel bearing of aircraft, 680 distance between two planes, 262 flight time and ticket price, 289 frequent flyer miles, 689 holding pattern, 633 parking at O’Hare International Airport, 245 revising a flight plan, 697 speed and direction of aircraft, 759, 763 Archaeology age of ancient tools, 480–481 age of fossil, 486 age of tree, 486 date of prehistoric man’s death, 500 Architecture brick staircase, 955, 979 Burj Khalifa building, 31 Flatiron Building, 703 floor design, 953–954, 979 football stadium seating, 955 mosaic design, 955, 979 Norman window, 37, 309 One World Trade Center, 541 parabolic arch, 309 racetrack design, 798 special window, 309, 317 stadium construction, 955 window design, 309 window dimensions, 102 xxvi Art fine decorative pieces, 541 framing a painting, 146 Astronomy angle of elevation of Sun, 679 distance from Earth to its moon, 29 distances of planets from Sun, 949 International Space Station (ISS), 838 light-year, 29 planetary orbits Earth, 798 elliptical, 798 Jupiter, 798 Mars, 798 Mercury, 825 Pluto, 798 Aviation modeling aircraft motion, 777 orbital launches, 856 Biology alcohol and driving, 447, 452 bacterial growth, 479–480, 493 E-coli, 235, 275 blood types, 987–988 bone length, 317–318 cricket chirp rate and temperature, 311 healing of wounds, 437, 451 maternal age versus Down syndrome, 290 muscle force, 764 yeast biomass as function of time, 492 Business advertising, 318 automobile production, 409, 872 blending coffee, 141 candy bar size, 103 checkout lines, 1006 clothing store, 1008 commissions, 317 cookie orders, 935 copying machines, 146 cost of can, 364, 367 of commodity, 410 of manufacturing, 29, 141, 374, 924–925 marginal, 301, 317 minimizing, 317, 931, 936 of printing, 337 of production, 234, 409, 899, 936 of transporting goods, 246 cost equation, 180, 192 cost function, 282 average, 217 demand for candy, 192 demand equation, 317, 400 depreciation, 402 discount pricing, 91, 92, 410 drive-thru rate at Burger King, 433 at Citibank, 437, 451 at McDonald’s, 438 equipment depreciation, 965 ethanol production, 493 expense computation, 142 farm workers in U.S., 487 Jiffy Lube’s car arrival rate, 437, 451 managing a meat market, 931 milk production, 494 mixing candy, 141 mixing nuts, 141 orange juice production, 872 precision ball bearings, 29 presale orders, 856 price markup, 91 of new car, 129 product design, 932 production scheduling, 931 product promotion, 181 profit, 899–900 maximizing, 929–930, 931–932 profit function, 213 rate of return on, 475 restaurant management, 856 revenue, 141, 301, 314–315 airline, 932 of clothing store, 889 daily, 301 from digital music, 259 from football seating, 966 maximizing, 301, 308 monthly, 301 theater, 857 revenue equation, 192 RV rental, 318–319 salary, 410, 955 gross, 212 increases in, 965, 979 Applications Index sales commission on, 128–129 of movie theater ticket, 844, 848–849, 856 net, 156 salvage value, 500 straight-line depreciation, 278–279, 282 supply and demand, 279–280, 282 tax, 374 theater attendance, 92 toy truck manufacturing, 924–925 transporting goods, 925 truck rentals, 180, 282–283 unemployment, 1009 wages of car salesperson, 180 hourly, 89, 91 Calculus absolute maximum/minimum, 228 area under a curve, 234, 260, 619 asymptotes, 346–347 average rate of change in, 230 bending wire into geometric shapes, 264 carrying a ladder around a corner, 633–634 cylinder inscribed in a cone, 264 cylinder inscribed in a sphere, 264 difference quotient, 205–206, 212, 439, 460, 654 express as single quotient in, 77 expressions with rational exponents in, 77 factoring in, 54, 58, 80 filling conical tank, 265 functions approximated by polynomial functions, 342 increasing/decreasing functions, 225–226 inequalities involving absolute value, 134 infinite geometric series, 960 infinite limits, 332 Intermediate Value Theorem, 384, 385 limit notation, 580, 582, 960 limits at infinity, 332 local maxima/minima, 227 longest ladder carried around corner, 586, 633 maximizing projectile range, 633, 659, 664 maximizing rain gutter construction, 663–664 open box construction, 265 partial fraction decomposition, 902 quadratic equations, 99–100, 102 radians, 509 reducing expression to lowest terms, 72 secant line, 231 Simpson’s rule, 309 Snell’s Law of Refraction, 634–635 tangent line, 703 the number e, 431–432 trigonometric functions, 657–658, 665 Carpentry. See also Construction pitch, 182 Chemistry, 91 alpha particles, 811 decomposition reactions, 487 drug concentration, 366 gas laws, 193 mixing acids, 146 pH, 450 purity of gold, 142 radioactive decay, 486, 493–494, 500, 932 radioactivity from Chernobyl, 487 reactions, 309 salt solutions, 142, 146 sugar molecules, 142 volume of gas, 128 Combinatorics airport codes, 989 binary codes, 1008 birthday permutations, 991, 995, 996, 1002–1003, 1007, 1008–1009 blouses and skirts combinations, 987 book arrangements, 995 box stacking, 995 code formation, 995 combination locks, 996 committee formation, 993, 995–996, 1008 Senate committees, 996 flag arrangement, 994, 1008 gender composition of children in family, 1000 letter codes, 989 license plate possibilities, 995, 1008, 1009 lining up people, 990, 995 number formation, 987, 995, 996, 1009 objects selection, 996 seating arrangements, 1008 shirts and ties combinations, 987 telephone numbers, 1008 two-symbol codewords, 986 word formation, 993–994, 996, 1009 Communications cell phone towers, 495 installing cable TV, 265 international call plan, 283 phone charges, 282 satellite dish, 785–786, 787 spreading of rumors, 437, 451 surveillance satellites, 681 tablet service, 245 Touch-Tone phones, 668, 713 wireless data plan, 198, 234–235, 271–272 Computers and computing comparing tablets, 103 graphics, 765, 900–901 households owning computers, 487 Internet searches, 112 iPod storage capacity, 283 xxvii laser printers, 142 three-click rule, 900 website design, 900 website map, 900 Word users, 487 Construction of border around a garden, 103 of border around a pool, 103 of box, 99–100, 102, 916 closed, 269 open, 265 of brick staircase, 979 of can, 398 of coffee can, 143 of cylindrical tube, 916 of enclosures around garden, 142 around pond, 142 maximizing area of, 304, 308, 317 of fencing, 304, 308, 317, 916 minimum cost for, 366 of flashlight, 787 of headlight, 787 of highway, 540, 690, 715 installing cable TV, 265 painting a room, 586 patio dimensions, 103 pitch of roof, 680 of rain gutter, 309, 534–535, 663–664 of ramp, 689 access ramp, 181 of rectangular field enclosure, 308 of stadium, 309, 955 of steel drum, 367 of swimming pool, 37, 38 of swing set, 698 of tent, 702 TV dish, 787 vent pipe installation, 798 Cryptography matrices in, 900 Decorating Christmas tree, 32 Demographics birth rate age of mother and, 311 of unmarried women, 301 diversity index, 450 divorced population, 306–307 marital status, 988 mosquito colony growth, 486 population. See Population rabbit colony growth, 948 Design of awning, 691 of box with minimum surface area, 367 of fine decorative pieces, 541 xxviii Applications Index of Little League Field, 518–519 of water sprinkler, 516 Direction of aircraft, 759, 763 compass heading, 764 for crossing a river, 763 of fireworks display, 810 of lightning strikes, 810 of motorboat, 763 of swimmer, 775 Distance Bermuda Triangle, 38 bicycle riding, 222 from Chicago to Honolulu, 619 circumference of Earth, 518 between cities, 511–512, 517 between Earth and Mercury, 691 between Earth and Venus, 691 from Earth to a star, 680 of explosion, 811 height of aircraft, 689, 691 of bouncing ball, 965, 979 of bridge, 689 of building, 680 of cloud, 536–537 of CN Tower, 540 of Eiffel Tower, 540 of embankment, 680 of Ferris Wheel rider, 633 of Great Pyramid of Cheops, 38, 691 of helicopter, 715 of hot-air balloon, 540 of Lincoln’s caricature on Mt. Rushmore, 540 of mountain, 686, 689 of Mt. Everest, 29 of One World Trade Center, 541 of statue on a building, 537 of tower, 540, 541 of tree, 689 of Washington Monument, 540 of Willis Tower, 680 from home, 222 from Honolulu to Melbourne, Australia, 619 of hot-air balloon to airport, 716 from intersection, 156 from intersection, 264 length of guy wire, 540, 542, 697 of lake, 602 of ski lift, 689 limiting magnitude of telescope, 500 to the Moon, 690 nautical miles, 518 pendulum swings, 961, 965 to plateau, 540 across a pond, 540 range of airplane, 143 reach of ladder, 540 of rotating beacon, 586 between runners, 689 at sea, 690 of search and rescue, 146 to shore, 540, 602, 690 between skyscrapers, 680 sound to measure, 118–119 stopping, 213, 301, 422 of storm, 145 to tower, 691 traveled by wheel, 37 between two moving vehicles, 156 toward intersection, 264 between two objects, 540 between two planes, 262 visibility of Gibb’s Hill Lighthouse beam, 38, 677–678, 681 visual, 38 walking, 222 width of gorge, 539 of Mississippi River, 680 of river, 536, 602 Economics Consumer Price Index (CPI), 477 demand equations, 400 federal stimulus package of 2009, 476 inflation, 476 IS-LM model in, 857 marginal propensity to consume, 966 multiplier, 966 national debt, 235 participation rate, 213 per capita federal debt, 476 poverty rates, 341 poverty threshold, 157 relative income of child, 900 unemployment, 1009 Education age distribution of community college, 1009 college costs, 476, 965–966 college tuition and fees, 899 computing grades, 129 degrees awarded, 985 doctorates, 1006 faculty composition, 1007 field trip, 374 fraternity purchase, 103 funding a college education, 500 GPA and work relationship, 103 grades, 91 learning curve, 438, 451 maximum level achieved, 937–938 median earnings and level of, 103 multiple-choice test, 995 probability of acceptance to college, 1009 spring break, 931 student loan, 270 interest on, 899 true/false test, 995 video games and grade-point average, 289 Electricity, 91 alternating current (ac), 602, 654 alternating current (ac) circuits, 577, 595 alternating current (ac) generators, 577–578 charging a capacitor, 713 cost of, 243 current in RC circuit, 438 current in RL circuit, 438, 451 impedance, 112 Kirchhoff’s Rules, 857, 873 Ohm’s law, 126 parallel circuits, 112 resistance in, 352 rates for, 129, 180 resistance, 70, 72, 193, 196, 352 voltage foreign, 29 household, 133 U.S., 29 Electronics. See also Computers and computing Blu-ray drive, 516 DVD drive, 516 loudspeakers, 712 microphones, 166 sawtooth curve, 664, 713 Energy nuclear power plant, 810–811 solar, 166, 771–772 solar heat, 788 thermostat control, 259 Engineering bridges clearance, 578 Golden Gate, 305–306 parabolic arch, 317, 788 semielliptical arch, 797, 798, 840 suspension, 309, 787–788 crushing load, 119 drive wheels of engine, 681 electrical, 529 Gateway Arch (St. Louis), 788 grade of mountain trail, 917 of road, 182 horsepower, 193 lean of Leaning Tower of Pisa, 690 maximum weight supportable by pine, 190 moment of inertia, 668 piston engines, 539 product of inertia, 664 road system, 728 rods and pistons, 698 safe load for a beam, 193 Applications Index searchlight, 642, 788, 840 whispering galleries, 797 Entertainment Demon Roller Coaster customer rate, 438 movie theater, 618 theater revenues, 857 Environment endangered species, 437 lake pollution control laws, 948 oil leakage, 409 Finance, 91. See also Investment(s) balancing a checkbook, 29 bills in wallet, 1009 cable rates, 494 clothes shopping, 937 college costs, 476, 965–966 computer system purchase, 475 cost of car, 92, 180 of car rental, 246 of electricity, 243 of fast food, 856 of land, 715 minimizing, 317, 366 of natural gas, 245 of pizza, 92 of printing, 337 of trans-Atlantic travel, 212–213, 221 of triangular lot, 702 cost equation, 192 cost function, 282 cost minimization, 301 credit cards balance on, 909 debt, 948 interest on, 475 payment, 246, 948 depreciation, 437 of car, 467, 503 discounts, 410 division of money, 88, 91 effective rate of interest, 472 electricity rates, 180 federal stimulus package of 2009, 476 financial planning, 136–137, 856, 869–870, 872 foreign exchange, 410 fraternity purchase, 103 funding a college education, 500 future value of money, 341–342 gross salary, 212 income versus crime rate, 496 inheritance, 146 life cycle hypothesis, 310 loans, 141 car, 948 interest on, 81, 136, 145, 147–148, 270, 899 repayment of, 475 student, 899 median earnings and level of education, 103 mortgages, 477 fees, 246 interest rates on, 476 payments, 189, 192, 196 second, 476 price appreciation of homes, 475 prices of fast food, 858 refunds, 856 revenue equation, 192 revenue maximization, 301, 302–304, 308 rich man’s promise, 966 salary options, 966 sales commission, 128–129 saving for a car, 475 for a home, 965 savings accounts interest, 475 selling price of a home, 197 sinking fund, 965–966 taxes, 282 e-filing returns, 235 federal income, 246, 410, 422 luxury, 282 truck rentals, 281 used-car purchase, 475 water bills, 129 Food and nutrition animal, 932 calories, 92 candy, 288 color mix of candy, 1009 cooler contents, 1009 cooling time of pizza, 486 fast food, 856, 858 Girl Scout cookies, 1006 hospital diet, 857, 872 ice cream, 931 “light” foods, 129 number of possible meals, 985–986 raisins, 288–289 soda and hot dogs buying combinations, 283 warming time of beer stein, 487 Forestry wood product classification, 485 Games coin toss, 999 die rolling, 998–999, 1000, 1009 grains of wheat on a chess board, 966 lottery, 1009, 1010–1011 Gardens and gardening. See also Landscaping border around, 103 enclosure for, 142 Geography area of Bermuda Triangle, 703 area of lake, 703, 715 inclination of mountain trail, 677, 715 Geology earthquakes, 452 Geometry angle between two lines, 654 balloon volume, 409 box volume, 773 circle area of, 141, 703 center of, 188 circumference of, 28, 141 equation of, 883 inscribed in square, 263 length of chord of, 698 radius of, 188, 916 collinear points, 883 cone volume, 193, 410 cube length of edge of, 389 surface area of, 29 volume of, 29 cylinder inscribing in cone, 264 inscribing in sphere, 264 volume of, 193, 410 Descartes’s method of equal roots, 916–917 equation of line, 883 ladder angle, 716 polygon area of, 883 diagonals of, 103 Pythagorean Theorem, 102 quadrilateral area, 717 rectangle area of, 28, 212, 261–262, 269 dimensions of, 92, 102, 146, 916 inscribed in circle, 263 inscribed in ellipse, 798 inscribed in semicircle, 263, 664 perimeter of, 28 semicircle inscribed in, 264 semicircle area, 702, 703, 717 sphere surface area of, 28 volume of, 28 square area of, 37, 141 diagonals of, 156 perimeter of, 141 surface area of balloon, 409 of cube, 29 of sphere, 28 triangle area of, 28, 702, 703, 717, 883 circumscribing, 692 equilateral, 28, 156 inscribed in circle, 264 isosceles, 212, 530, 717, 916 medians of, 155 Pascal’s, 948 perimeter of, 28 xxix xxx Applications Index triangle (continued ) right, 539, 679 sides of, 717 Government federal debt, 235 per capita, 476 federal income tax, 213, 246, 410, 422 e-filing returns, 235 federal stimulus package of 2009, 476 federal tax withholding, 129 first-class mail, 247 Health. See also Medicine Landscaping. See also Gardens and gardening height of tree, 689 pond enclosure, 317 rectangular pond border, 317 removing stump, 765 tree planting, 872 watering lawn, 516 Law and law enforcement income vs. crime rate, 496 motor vehicle thefts, 1006 violent crimes, 213 volume of balloon, 409 wire enclosure area, 264 Mixtures. See also Chemistry blending coffees, 137–138, 141, 147, 925, 935 blending teas, 141 cement, 143 mixed nuts, 141, 856, 925, 935 mixing candy, 141 solutions, 856 water and antifreeze, 142 Motion, 713. See also Physics age versus total cholesterol, 495 blood pressure, 633 cigarette use among teens, 181 exercising, 129 expenditures on, 213 heartbeats during exercise, 276–277 ideal body weight, 422 life cycle hypothesis, 310 life expectancy, 128 Leisure and recreation Home improvement. See also Construction Measurement catching a train, 840 on a circle, 516 of Ferris Wheel rider, 633 of golf ball, 220–221 minute hand of clock, 516, 602 objects approaching intersection, 836–837 of pendulum, 713 revolutions of circular disk, 37 simulating, 831 tortoise and the hare race, 916 uniform, 138–139, 141, 836–837 painting a house, 858 optical methods of, 642 of rainfall, 772 Motor vehicles Housing Mechanics, 91. See also Physics apartment rental, 310 number of rooms in, 212 price appreciation of homes, 475 Investment(s), 88, 91, 141, 145 annuity, 962–963, 965 in bonds, 932 Treasuries, 872, 873, 922, 924, 926 zero-coupon, 473, 476 in CDs, 472, 932 compound interest on, 468–469, 470, 471–472, 475–476 diversified, 858 dividing, 247 doubling of, 473–474, 476 effective rate of interest, 472 finance charges, 475 in fixed-income securities, 476, 932 401K, 965, 979 growth rate for, 475–476 IRA, 476, 962–963, 965 mutual fund growth over time, 490 return on, 475, 931, 932 savings account, 471–472 in stock analyzing, 320 appreciation, 475 beta, 273, 320 NASDAQ stocks, 995 NYSE stocks, 995 portfolios of, 988 price of, 966 time to reach goal, 475, 477 tripling of, 474, 476 amusement park ride, 516 cable TV, 265 rates, 494 community skating rink, 270 Ferris wheel, 187, 633, 691, 712 field trip, 374 video games and grade-point average, 289 Medicine. See also Health age versus total cholesterol, 495 blood pressure, 633 cancer breast, 493 pancreatic, 437 drug concentration, 234, 366 drug medication, 437, 451 healing of wounds, 437, 451 spreading of disease, 501 Meteorology weather balloon height and atmospheric pressure, 491 Miscellaneous banquet seating, 931 bending wire, 916 biorhythms, 578 carrying a ladder around a corner, 529, 586, 633–634 citrus ladders, 955 coffee container, 504 cross-sectional area of beam, 213, 220 curve fitting, 857, 872, 935 diameter of copper wire, 29 drafting error, 156 land dimensions, 689 Mandelbrot sets, 751 motor, 29 pet ownership, 1006 reading books, 133 surface area of balloon, 409 alcohol and driving, 447, 452 angular speed of race car, 602 approaching intersection, 836–837 automobile production, 409, 872 average car speed, 143 brake repair with tune-up, 1009 braking load, 772, 775 crankshafts, 690 depreciation, 402 depreciation of, 467, 503 with Global Positioning System (GPS), 501 loans for, 948 markup of new car, 129 runaway car, 315 speed and miles per gallon, 310–311 spin balancing tires, 517 stopping distance, 213, 301, 422 theft of, 1006 used-car purchase, 475 windshield wiper, 516 Music revenues from, 259 Navigation avoiding a tropical storm, 697 bearing, 678, 696 of aircraft, 680 of ship, 680, 715 charting a course, 764 commercial, 689 compass heading, 764 crossing a river, 763, 764 error in correcting, 694–695, 715 time lost due to, 689 Applications Index rescue at sea, 686–687, 689–690 revising a flight plan, 697 Oceanography tides, 596 Optics angle of refraction, 634–635 bending light, 635 Brewster angle, 635 index of refraction, 634–635 intensity of light, 193 laser beam, 679 laser projection, 664 lensmaker’s equation, 72 light obliterated through glass, 437 mirrors, 811 reflecting telescope, 788 Pediatrics height vs. head circumference, 289, 422 Pharmacy vitamin intake, 857, 873 Photography camera distance, 541 Physics, 91 angle of elevation of Sun, 679 angle of inclination, 772 bouncing balls, 979 braking load, 772 damped motion, 708, 716 diameter of atom, 29 Doppler effect, 366 effect of elevation on weight, 221 falling objects, 192 force, 141, 763 of attraction between two bodies, 192 to hold a wagon on a hill, 769–770 muscle, 764 resultant, 763 of wind on a window, 191, 193 gravity, 352, 374 on Earth, 212, 422 on Jupiter, 212 harmonic motion, 707 heat loss, 190, 196 heat transfer, 633 horsepower, 193 inclination of mountain trail, 677 inclined ramp, 764 intensity of light, 146, 193 kinetic energy, 141, 193 maximum weight supportable by pine, 190 missile trajectory, 320 moment of inertia, 668 motion of object, 707 Newton’s law, 192 Ohm’s law, 126 pendulum motion, 119, 516, 713, 961 period, 259–260, 422 simple pendulum, 192 pressure, 141, 192 product of inertia, 664 projectile motion, 102–103, 304–305, 308–309, 539, 552, 633, 634, 659, 664, 668, 759, 829–830, 836, 837, 840 artillery, 315, 624 hit object, 836 thrown object, 836 safe load for a beam, 193 simple harmonic motion, 716 simulating motion, 831 sound to measure distance, 118–119 speed of sound, 133 static equilibrium, 760–761, 764, 765, 775 static friction, 764 stress of materials, 193 stretching a spring, 192 tension, 760–761, 764, 775, 776, 971 thrown object, 146, 759 ball, 310, 314 truck pulls, 765 uniform motion, 141, 146, 264, 836–837, 840 velocity down inclined planes, 80 vertically propelled object, 314 vibrating string, 192 wavelength of visible light, 29 weight, 193, 196 of a boat, 763 of a car, 763 of a piano, 760 work, 141 Psychometrics IQ tests, 129 Pyrotechnics fireworks display, 810 Rate. See also Speed of car, 517 catching a bus, 836 catching a train, 836 current of stream, 857 of emptying fuel tanks, 146 oil tankers, 143 a pool, 143 a tub, 143 to keep up with the Sun, 518 revolutions per minute of bicycle wheels, 517 of pulleys, 519 speed average, 143 of current, 141 of cyclist, 143 of motorboat, 141 of moving walkways, 141–142 per gallon rate and, 310–311 of plane, 143 of sound, 133 of water use, 260 Real estate Play commission, 128–129 cost of land, 715 cost of triangular lot, 702 housing prices, 398 mortgage loans, 477 swinging, 717 wagon pulling, 763, 770 Recreation Population. See also Demographics bacteria, 439, 486, 493 decline in, 486 E-coli growth, 235, 275 of endangered species, 487 of fruit fly, 484 as function of age, 212 growth in, 486 insect, 352, 486 of trout, 948 of United States, 467, 494, 981 of world, 467, 495, 500, 939 Probability of birthday shared by people in a room, 487 checkout lines, 1006 classroom composition, 1006 exponential, 433, 437–438, 451 household annual income, 1006 Poisson, 438 “Price is Right” games, 1006 of winning a lottery, 1007 xxxi bungee jumping, 374 Demon Roller Coaster customer rate, 438 online gambling, 1006 Security security cameras, 680 Seismology calibrating instruments, 840 Sequences. See also Combinatorics ceramic tile floor design, 953–954 Drury Lane Theater, 955 football stadium seating, 955 seats in amphitheater, 955 Speed of aircraft, 763 angular, 517, 602 of current, 517, 935 as function of time, 222, 264 linear, 514 on Earth, 517 of Moon, 517 xxxii Applications Index revolutions per minute of pulley, 517 of rotation of lighthouse beacons, 602 of swimmer, 775 of truck, 679 of wheel pulling cable cars, 517 wind, 856 Sports baseball, 836, 837, 996, 1008 diamond, 156 dimensions of home plate, 702 field, 697, 698 Little League, 156, 518–519 on-base percentage, 284–285 stadium, 697 World Series, 996 basketball, 996 free throws, 220, 680–681 granny shots, 220 biathlon, 143 bungee jumping, 374 calculating pool shots, 541 distance between runners, 689 exacta betting, 1009 football, 142, 798 defensive squad, 996 field design, 103 seating revenue, 966 golf, 220–221, 496, 829–830, 836 distance to the green, 696 sand bunkers, 624 hammer throw, 603 Olympic heroes, 143 races, 142, 146, 913–914, 916 relay runners, 1008 swimming, 717, 775 tennis, 142 Statistics. See Probability Surveys Travel. See also Air travel; Navigation of appliance purchases, 987 data analysis, 984, 987 stock portfolios, 988 of summer session attendance, 987 of TV sets in a house, 1006 bearing, 715 drivers stopped by the police, 503 driving to school, 192 parking at O’Hare International Airport, 245 Temperature Volume of air parcel, 955 body, 29, 133 conversion of, 410, 422 cooling time of pizza, 486 cricket chirp rate and, 311 Fahrenheit from Celsius conversion, 87 measuring, 180–181 after midnight, 341 monthly, 595–596, 602 relationship between scales, 259 sinusoidal function from, 591–592 of skillet, 500 warming time of beer stein, 487 wind chill factor, 501 of gasoline in tank, 80 of ice in skating rink, 270 of water in cone, 265 Time for beer stein to warm, 487 for block to slide down inclined plane, 539 Ferris Wheel rider height as function of, 633 to go from an island to a town, 265 hours of daylight, 400–401, 505, 593–594, 597, 604, 618 for pizza to cool, 486 for rescue at sea, 146 of sunrise, 518, 618 of trip, 529, 541 Transportation deicing salt, 624 Niagara Falls Incline Railway, 680 Weapons artillery, 315, 624 cannons, 320 Weather atmospheric pressure, 437, 451 avoiding a tropical storm, 697 cooling air, 955 hurricanes, 341, 595 lightning and thunder, 145 lightning strikes, 807–808, 810 probability of rain, 1002 rainfall measurement, 772 relative humidity, 438 weather satellites, 187 wind chill, 246–247, 501 Work, 770 computing, 770, 771, 775 constant rate jobs, 936 GPA and, 103 pulling a wagon, 770 ramp angle, 772 wheelbarrow push, 763 working together, 140, 142, 146 R Review A Look Ahead Outline Chapter R, as the title states, contains review material. Your instructor may choose to cover all or part of it as a regular chapter at the beginning of your course or later as a just-in-time review when the content is required. Regardless, when information in this chapter is needed, a specific reference to this chapter will be made so you can review. R.1 R.2 R.3 R.4 R.5 R.6 R.7 R.8 Real Numbers Algebra Essentials Geometry Essentials Polynomials Factoring Polynomials Synthetic Division Rational Expressions nth Roots; Rational Exponents 1 1 2 CHAPTER R Review R.1 Real Numbers PREPARING FOR THIS TEXT Before getting started, read “To the Student” at the front of this text. OBJECTIVES 1 2 3 4 Work with Sets (p. 2) Classify Numbers (p. 4) Evaluate Numerical Expressions (p. 8) Work with Properties of Real Numbers (p. 9) 1 Work with Sets A set is a well-defined collection of distinct objects. The objects of a set are called its elements. By well-defined, we mean that there is a rule that enables us to determine whether a given object is an element of the set. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol ∅. For example, the set of digits consists of the collection of numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. If we use the symbol D to denote the set of digits, then we can write D = 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 96 In this notation, the braces 5 6 are used to enclose the objects, or elements, in the set. This method of denoting a set is called the roster method. A second way to denote a set is to use set-builder notation, where the set D of digits is written as D = { x 0 x is a digit } c c c 6 $%+++& Read as “D is the set of all x such that x is a digit.” 2 EX AMPLE 1 Using Set-builder Notation and the Roster Method (a) E = 5 x 0 x is an even digit6 = 5 0, 2, 4, 6, 86 (b) O = 5 x 0 x is an odd digit6 = 5 1, 3, 5, 7, 96 r Because the elements of a set are distinct, we never repeat elements. For example, we would never write 5 1, 2, 3, 26 ; the correct listing is 5 1, 2, 36 . Because a set is a collection, the order in which the elements are listed is immaterial. 5 1, 2, 36 , 5 1, 3, 26 , 5 2, 1, 36 , and so on, all represent the same set. If every element of a set A is also an element of a set B, then A is a subset of B, which is denoted A ⊆ B. If two sets A and B have the same elements, then A equals B, which is denoted A = B. For example, 5 1, 2, 36 ⊆ 5 1, 2, 3, 4, 56 and 5 1, 2, 36 = 5 2, 3, 16 . DEFINITION EX AMPLE 2 If A and B are sets, the intersection of A with B, denoted A ∩ B, is the set consisting of elements that belong to both A and B. The union of A with B, denoted A ∪ B, is the set consisting of elements that belong to either A or B, or both. Finding the Intersection and Union of Sets Let A = 5 1, 3, 5, 86 , B = 5 3, 5, 76 , and C = 5 2, 4, 6, 86 . Find: (a) A ∩ B (b) A ∪ B (c) B ∩ 1A ∪ C2 SECTION R.1 Real Numbers Solution (a) A ∩ B = 5 1, 3, 5, 86 ∩ 5 3, 5, 76 = 5 3, 56 (b) A ∪ B = 5 1, 3, 5, 86 ∪ 5 3, 5, 76 = 5 1, 3, 5, 7, 86 (c) B ∩ 1A ∪ C2 = 5 3, 5, 76 ∩ 3 5 1, 3, 5, 86 ∪ 5 2, 4, 6, 86 4 = 5 3, 5, 76 ∩ 5 1, 2, 3, 4, 5, 6, 86 = 5 3, 56 Now Work PROBLEM 3 r 15 Usually, in working with sets, we designate a universal set U, the set consisting of all the elements that we wish to consider. Once a universal set has been designated, we can consider elements of the universal set not found in a given set. DEFINITION EXAMPL E 3 If A is a set, the complement of A, denoted A, is the set consisting of all the elements in the universal set that are not in A.* Finding the Complement of a Set If the universal set is U = 5 1, 2, 3, 4, 5, 6, 7, 8, 96 and if A = 5 1, 3, 5, 7, 96 , then A = 5 2, 4, 6, 86 . r It follows from the definition of complement that A ∪ A = U and A ∩ A = ∅. Do you see why? Now Work Universal set B A PROBLEM 19 It is often helpful to draw pictures of sets. Such pictures, called Venn diagrams, represent sets as circles enclosed in a rectangle, which represents the universal set. Such diagrams often help us to visualize various relationships among sets. See Figure 1. If we know that A ⊆ B, we might use the Venn diagram in Figure 2(a). If we know that A and B have no elements in common—that is, if A ∩ B = ∅—we might use the Venn diagram in Figure 2(b). The sets A and B in Figure 2(b) are said to be disjoint. C Figure 1 Venn diagram Universal set Universal set B A A (a) A 債 B subset Figure 2 B (b) A 傽 B 5 ⭋ disjoint sets Figures 3(a), 3(b), and 3(c) use Venn diagrams to illustrate the definitions of intersection, union, and complement, respectively. Universal set A Figure 3 B (a) A B intersection Universal set Universal set A B (b) A B union *Some books use the notation A′ for the complement of A. A A (c) A complement 4 CHAPTER R Review 2 Classify Numbers It is helpful to classify the various kinds of numbers that we deal with as sets. The counting numbers, or natural numbers, are the numbers in the set 5 1, 2, 3, 4, c6 . (The three dots, called an ellipsis, indicate that the pattern continues indefinitely.) As their name implies, these numbers are often used to count things. For example, there are 26 letters in our alphabet; there are 100 cents in a dollar. The whole numbers are the numbers in the set 5 0, 1, 2, 3, c6 —that is, the counting numbers together with 0. The set of counting numbers is a subset of the set of whole numbers. DEFINITION The integers are the set of numbers 5 c, - 3, - 2, - 1, 0, 1, 2, 3,c6 . These numbers are useful in many situations. For example, if your checking account has $10 in it and you write a check for $15, you can represent the current balance as −$5. Each time we expand a number system, such as from the whole numbers to the integers, we do so in order to be able to handle new, and usually more complicated, problems. The integers enable us to solve problems requiring both positive and negative counting numbers, such as profit/loss, height above/below sea level, temperature above/below 0°F, and so on. But integers alone are not sufficient for all problems. For example, they do not answer the question “What part of a dollar is 38 cents?” To answer such a question, 38 we enlarge our number system to include rational numbers. For example, 100 answers the question “What part of a dollar is 38 cents?” DEFINITION a of two b integers. The integer a is called the numerator, and the integer b, which cannot be 0, is called the denominator. The rational numbers are the numbers in the a set e x ` x = , where a, b are integers and b ≠ 0 f. b A rational number is a number that can be expressed as a quotient 3 5 0 2 100 a , , , - , and . Since = a for any 4 2 4 3 3 1 integer a, it follows that the set of integers is a subset of the set of rational numbers. Rational numbers may be represented as decimals. For example, the rational 2 7 3 5 may be represented as decimals by merely carrying out numbers , , - , and 4 2 3 66 the indicated division: Examples of rational numbers are 3 = 0.75 4 5 = 2.5 2 - 2 = - 0.666c = - 0.6 3 7 = 0.1060606c = 0.106 66 5 3 Notice that the decimal representations of and terminate, or end. The decimal 4 2 7 2 representations of - and do not terminate, but they do exhibit a pattern of 66 2 3 repetition. For - , the 6 repeats indefinitely, as indicated by the bar over the 6; for 3 7 , the block 06 repeats indefinitely, as indicated by the bar over the 06. It can be 66 shown that every rational number may be represented by a decimal that either terminates or is nonterminating with a repeating block of digits, and vice versa. On the other hand, some decimals do not fit into either of these categories. Such decimals represent irrational numbers. Every irrational number may be represented by a decimal that neither repeats nor terminates. In other words, irrational numbers a cannot be written in the form , where a, b are integers and b ≠ 0. b SECTION R.1 Real Numbers 5 Irrational numbers occur naturally. For example, consider the isosceles right triangle whose legs are each of length 1. See Figure 4. The length of the hypotenuse is 12, an irrational number. Also, the number that equals the ratio of the circumference C to the diameter d of any circle, denoted by the symbol p (the Greek letter pi), is an irrational number. See Figure 5. C 2 1 d 1 Figure 5 p = Figure 4 DEFINITION C d The set of real numbers is the union of the set of rational numbers with the set of irrational numbers. Figure 6 shows the relationship of various types of numbers.* Irrational numbers Rational numbers Integers Whole numbers Natural or counting numbers Real numbers Figure 6 EXAMPL E 4 Classifying the Numbers in a Set List the numbers in the set 4 b - 3, , 0.12, 22, p, 10, 2.151515c1where the block 15 repeats2 r 3 that are (a) Natural numbers (d) Irrational numbers Solution (b) Integers (e) Real numbers (c) Rational numbers (a) 10 is the only natural number. (b) - 3 and 10 are integers. 4 (c) - 3, 10, , 0.12, and 2.151515c are rational numbers. 3 (d) 12 and p are irrational numbers. (e) All the numbers listed are real numbers. Now Work PROBLEM 25 *The set of real numbers is a subset of the set of complex numbers. We discuss complex numbers in Chapter 1, Section 1.3. r 6 CHAPTER R Review Approximations Every decimal may be represented by a real number (either rational or irrational), and every real number may be represented by a decimal. In practice, the decimal representation of an irrational number is given as an approximation. For example, using the symbol ≈ (read as “approximately equal to”), we can write 22 ≈ 1.4142 p ≈ 3.1416 In approximating decimals, we either round off or truncate to a given number of decimal places.* The number of places establishes the location of the final digit in the decimal approximation. Truncation: Drop all of the digits that follow the specified final digit in the decimal. Rounding: Identify the specified final digit in the decimal. If the next digit is 5 or more, add 1 to the final digit; if the next digit is 4 or less, leave the final digit as it is. Then truncate following the final digit. EX AMPLE 5 Approximating a Decimal to Two Places Approximate 20.98752 to two decimal places by (a) Truncating (b) Rounding Solution For 20.98752, the final digit is 8, since it is two decimal places from the decimal point. (a) To truncate, we remove all digits following the final digit 8. The truncation of 20.98752 to two decimal places is 20.98. (b) The digit following the final digit 8 is the digit 7. Since 7 is 5 or more, we add 1 to the final digit 8 and truncate. The rounded form of 20.98752 to two decimal places is 20.99. r EX AMPLE 6 Approximating a Decimal to Two and Four Places Rounded to Two Decimal Places Number Rounded to Four Decimal Places Truncated to Two Decimal Places Truncated to Four Decimal Places (a) 3.14159 3.14 3.1416 3.14 3.1415 (b) 0.056128 0.06 0.0561 0.05 0.0561 893.46 893.4613 893.46 893.4612 (c) 893.46125 Now Work PROBLEM r 29 Calculators Calculators are incapable of displaying decimals that contain a large number of digits. For example, some calculators are capable of displaying only eight digits. When a number requires more than eight digits, the calculator either truncates or rounds. * Sometimes we say “correct to a given number of decimal places” instead of “truncate.” SECTION R.1 Real Numbers 7 To see how your calculator handles decimals, divide 2 by 3. How many digits do you see? Is the last digit a 6 or a 7? If it is a 6, your calculator truncates; if it is a 7, your calculator rounds. There are different kinds of calculators. An arithmetic calculator can only add, subtract, multiply, and divide numbers; therefore, this type is not adequate for this course. Scientific calculators have all the capabilities of arithmetic calculators and also contain function keys labeled ln, log, sin, cos, tan, xy, inv, and so on. As you proceed through this text, you will discover how to use many of the function keys. Graphing calculators have all the capabilities of scientific calculators and contain a screen on which graphs can be displayed. For those who have access to a graphing calculator, we have included comments, examples, and exercises marked with a , indicating that a graphing calculator is required. We have also included an appendix that explains some of the capabilities of a graphing calculator. The comments, examples, and exercises may be omitted without loss of continuity, if so desired. Operations In algebra, we use letters such as x, y, a, b, and c to represent numbers. The symbols used in algebra for the operations of addition, subtraction, multiplication, and division are +, - , # , and >. The words used to describe the results of these operations are sum, difference, product, and quotient. Table 1 summarizes these ideas. Table 1 Operation Symbol Words Addition a + b Sum: a plus b Subtraction a - b Difference: a minus b Multiplication a # b, (a) # b, a # (b), (a) # (b), ab, (a)b, a(b), (a)(b) Product: a times b Division a>b or a b Quotient: a divided by b In algebra, we generally avoid using the multiplication sign * and the division sign , so familiar in arithmetic. Notice also that when two expressions are placed next to each other without an operation symbol, as in ab, or in parentheses, as in 1a2 1b2, it is understood that the expressions, called factors, are to be multiplied. We also prefer not to use mixed numbers in algebra. When mixed numbers 3 3 are used, addition is understood; for example, 2 means 2 + . In algebra, use of 4 4 a mixed number may be confusing because the absence of an operation symbol 3 between two terms is generally taken to mean multiplication. The expression 2 is 4 11 therefore written instead as 2.75 or as . 4 The symbol =, called an equal sign and read as “equals” or “is,” is used to express the idea that the number or expression on the left of the equal sign is equivalent to the number or expression on the right. EXAMPL E 7 Writing Statements Using Symbols (a) The sum of 2 and 7 equals 9. In symbols, this statement is written as 2 + 7 = 9. (b) The product of 3 and 5 is 15. In symbols, this statement is written as 3 # 5 = 15. Now Work PROBLEM 41 r 8 CHAPTER R Review 3 Evaluate Numerical Expressions Consider the expression 2 + 3 # 6. It is not clear whether we should add 2 and 3 to get 5, and then multiply by 6 to get 30; or first multiply 3 and 6 to get 18, and then add 2 to get 20. To avoid this ambiguity, we have the following agreement. In Words Multiply first, then add. We agree that whenever the two operations of addition and multiplication separate three numbers, the multiplication operation will always be performed first, followed by the addition operation. For 2 + 3 # 6, then, we have 2 + 3 # 6 = 2 + 18 = 20 EX AMPLE 8 Finding the Value of an Expression Evaluate each expression. (a) 3 + 4 # 5 Solution (b) 8 # 2 + 1 (a) 3 + 4 # 5 = 3 + 20 = 23 (c) 2 + 2 # 2 (b) 8 # 2 + 1 = 16 + 1 = 17 c Multiply first c Multiply first r (c) 2 + 2 # 2 = 2 + 4 = 6 Now Work PROBLEM 53 When we want to indicate adding 3 and 4 and then multiplying the result by 5, we use parentheses and write 13 + 42 # 5. Whenever parentheses appear in an expression, it means “perform the operations within the parentheses first!” EX AMPLE 9 Finding the Value of an Expression (a) 15 + 32 # 4 = 8 # 4 = 32 (b) 14 + 52 # 18 - 22 = 9 # 6 = 54 r When we divide two expressions, as in 2 + 3 4 + 8 it is understood that the division bar acts like parentheses; that is, 12 + 32 2 + 3 = 4 + 8 14 + 82 Rules for the Order of Operations 1. Begin with the innermost parentheses and work outward. Remember that in dividing two expressions, we treat the numerator and denominator as if they were enclosed in parentheses. 2. Perform multiplications and divisions, working from left to right. 3. Perform additions and subtractions, working from left to right. SECTION R.1 Real Numbers EX AM PL E 1 0 Finding the Value of an Expression Evaluate each expression. (b) 5 # 13 + 42 + 2 (a) 8 # 2 + 3 2 + 5 (c) 2 + 4#7 Solution 9 (d) 2 + 3 4 + 2 # 110 + 62 4 (a) 8 # 2 + 3 = 16 + 3 = 19 c Multiply first (b) 5 # 13 + 42 + 2 = 5 # 7 + 2 = 35 + 2 = 37 c c Parentheses first Multiply before adding 2 + 5 2 + 5 7 = = # 2 + 4 7 2 + 28 30 (d) 2 + 3 4 + 2 # 110 + 62 4 = 2 + 3 4 + 2 # 1162 4 = 2 + 3 4 + 324 = 2 + 3 364 = 38 (c) r Be careful if you use a calculator. For Example 10(c), you need to use parentheses. See Figure 7.* If you don’t, the calculator will compute the expression 2 + 5 + 4 # 7 = 2 + 2.5 + 28 = 32.5 2 giving a wrong answer. Now Work Figure 7 PROBLEMS 59 AND 67 4 Work with Properties of Real Numbers The equal sign is used to mean that one expression is equivalent to another. Four important properties of equality are listed next. In this list, a, b, and c represent real numbers. 1. 2. 3. 4. The reflexive property states that a number equals itself; that is, a = a. The symmetric property states that if a = b, then b = a. The transitive property states that if a = b and b = c, then a = c. The principle of substitution states that if a = b, then we may substitute b for a in any expression containing a. Now, let’s consider some other properties of real numbers. EXAMPL E 11 Commutative Properties (a) 3 + 5 = 8 5 + 3 = 8 3 + 5 = 5 + 3 (b) 2 # 3 = 6 3#2 = 6 2#3 = 3#2 r This example illustrates the commutative property of real numbers, which states that the order in which addition or multiplication takes place does not affect the final result. * Notice that we converted the decimal to its fraction form. Another option, when using a TI-84 Plus C, is to use the fraction template under the MATH button to enter the expression as it appears in Example 10(c). Consult your manual to see how to enter such expressions on your calculator. 10 CHAPTER R Review Commutative Properties a + b = b + a (1a) a#b = b#a (1b) Here, and in the properties listed next and on pages 11–13, a, b, and c represent real numbers. EX A MPL E 1 2 Associative Properties (a) 2 + 13 + 42 = 2 + 7 = 9 12 + 32 + 4 = 5 + 4 = 9 2 + 13 + 42 = 12 + 32 + 4 (b) 2 # 13 # 42 = 2 # 12 = 24 12 # 32 # 4 = 6 # 4 = 24 2 # 13 # 42 = 12 # 32 # 4 r The way we add or multiply three real numbers does not affect the final result. Expressions such as 2 + 3 + 4 and 3 # 4 # 5 present no ambiguity, even though addition and multiplication are performed on one pair of numbers at a time. This property is called the associative property. Associative Properties a + 1b + c2 = 1a + b2 + c = a + b + c a # 1b # c2 = 1a # b2 # c = a # b # c (2a) (2b) Distributive Property a # 1b + c2 = a # b + a # c 1a + b2 # c = a # c + b # c (3a) (3b) The distributive property may be used in two different ways. EX A MPL E 1 3 Distributive Property (a) 2 # 1x + 32 = 2 # x + 2 # 3 = 2x + 6 Use to remove parentheses. Use to combine two expressions. (b) 3x + 5x = 13 + 52x = 8x (c) 1x + 22 1x + 32 = x1x + 32 + 21x + 32 = 1x2 + 3x2 + 12x + 62 = x2 + (3x + 2x) + 6 = x2 + 5x + 6 Now Work PROBLEM r 89 The real numbers 0 and 1 have unique properties called the identity properties. EX A MPL E 1 4 Identity Properties (a) 4 + 0 = 0 + 4 = 4 (b) 3 # 1 = 1 # 3 = 3 r SECTION R.1 Real Numbers 11 Identity Properties 0 + a = a + 0 = a (4a) a#1 = 1#a = a (4b) We call 0 the additive identity and 1 the multiplicative identity. For each real number a, there is a real number - a, called the additive inverse of a, having the following property: Additive Inverse Property a + 1 - a2 = - a + a = 0 EX AM PL E 15 (5a) Finding an Additive Inverse (a) The additive inverse of 6 is - 6, because 6 + 1 - 62 = 0. (b) The additive inverse of - 8 is - 1 - 82 = 8, because - 8 + 8 = 0. r The additive inverse of a, that is, - a, is often called the negative of a or the opposite of a. The use of such terms can be dangerous, because they suggest that the additive inverse is a negative number, which may not be the case. For example, the additive inverse of - 3, or - 1 - 32, equals 3, a positive number. 1 For each nonzero real number a, there is a real number , called the multiplicative a inverse of a, having the following property: Multiplicative Inverse Property a# 1 1 = #a = 1 a a The multiplicative inverse if a ≠ 0 (5b) 1 of a nonzero real number a is also referred to as the a reciprocal of a. EX AM PL E 1 6 Finding a Reciprocal 1 1 (a) The reciprocal of 6 is , because 6 # = 1. 6 6 1 1 (b) The reciprocal of - 3 is , because - 3 # = 1. -3 -3 2 3 2 3 (c) The reciprocal of is , because # = 1. 3 2 3 2 r With these properties for adding and multiplying real numbers, we can define the operations of subtraction and division as follows: DEFINITION The difference a - b, also read “a less b” or “a minus b,” is defined as a - b = a + 1 - b2 To subtract b from a, add the opposite of b to a. (6) 12 CHAPTER R Review DEFINITION a If b is a nonzero real number, the quotient , also read as “a divided by b” or b “the ratio of a to b,” is defined as a 1 = a# b b EX A MPL E 1 7 if b ≠ 0 (7) Working with Differences and Quotients (a) 8 - 5 = 8 + 1 - 52 = 3 (b) 4 - 9 = 4 + 1 - 92 = - 5 5 1 (c) = 5# 8 8 r For any number a, the product of a times 0 is always 0; that is, In Words Multiplication by Zero a#0 = 0 The result of multiplying by zero is zero. (8) For a nonzero number a, Division Properties 0 = 0 a a = 1 if a ≠ 0 a (9) 2 = x means to 0 find x such that 0 # x = 2. But 0 # x equals 0 for all x, so there is no unique number x such that 2 = x. ■ 0 NOTE Division by 0 is not defined. One reason is to avoid the following difficulty: Rules of Signs a1 - b2 = - 1ab2 - 1 - a2 = a EX A MPL E 1 8 Applying the Rules of Signs (a) 21 - 32 = - 12 # 32 = - 6 3 -3 3 (c) = = -2 2 2 x 1 # 1 (e) = x = - x -2 -2 2 1 - a2b = - 1ab2 a -a a = = -b b b 1 - a2 1 - b2 = ab -a a = (10) -b b (b) 1 - 32 1 - 52 = 3 # 5 = 15 -4 4 (d) = -9 9 r SECTION R.1 Real Numbers 13 Cancellation Properties ac = bc implies a = b if c ≠ 0 ac a = if b ≠ 0, c ≠ 0 bc b EX AM PL E 1 9 (11) Using the Cancellation Properties (a) If 2x = 6, then 2x = 6 2x = 2 # 3 x = 3 NOTE We follow the common practice of using slash marks to indicate cancellations. ■ In Words (b) 18 = 12 3# 6 2# 6 Factor 6. Cancel the 2’s. 3 2 c = r Cancel the 6’s. Zero-Product Property If a product equals 0, then one or both of the factors is 0. EX AM PL E 2 0 If ab = 0, then a = 0, or b = 0, or both. (12) Using the Zero-Product Property r If 2x = 0, then either 2 = 0 or x = 0. Since 2 ≠ 0, it follows that x = 0. Arithmetic of Quotients a c ad bc ad + bc + = + = b d bd bd bd a#c ac = b d bd a a d b ad = # = c b c bc d EX AM PL E 2 1 if b ≠ 0, d ≠ 0 (13) if b ≠ 0, d ≠ 0 (14) if b ≠ 0, c ≠ 0, d ≠ 0 (15) Adding, Subtracting, Multiplying, and Dividing Quotients (a) 2 5 2#2 3#5 2#2 + 3#5 4 + 15 19 + = # + # = = = # 3 2 3 2 3 2 3 2 6 6 c By equation (13) (b) 3 2 3 2 3 -2 - = + a- b = + 5 3 5 3 5 3 c By equation (6) = c By equation (10) 3 # 3 + 5 # 1 - 22 9 + 1 - 102 -1 1 = = = # 5 3 15 15 15 c By equation (13) 14 CHAPTER R Review NOTE Slanting the cancellation marks in different directions for different factors, as shown here, is a good practice to follow, since it will help in checking for errors. ■ (c) 2#5 8 # 15 8 # 15 2# 4 # 3 #5 = = 10 = # = # # 3 4 1 1 3 4 3 4 c c By equation (14) By equation (11) 3 5 3 9 3#9 27 (d) = # = # = 7 5 7 c 5 7 35 By equation (14) 9 æ r By equation (15) NOTE In writing quotients, we shall follow the usual convention and write the quotient in lowest terms. That is, we write it so that any common factors of the numerator and the denominator have been removed using the cancellation properties, equation (11). As examples, 90 15 # 6 15 = = 24 4# 6 4 4# 6 #x# x 4x 24x 2 = = 18x 3# 6 # x 3 Now Work PROBLEMS 69, 73, AND x ≠ 0 ■ 83 Sometimes it is easier to add two fractions using least common multiples (LCM). The LCM of two numbers is the smallest number that each has as a common multiple. EX A MPL E 2 2 Finding the Least Common Multiple of Two Numbers Find the least common multiple of 15 and 12. Solution To find the LCM of 15 and 12, we look at multiples of 15 and 12. 15, 30, 45, 60, 75, 90, 105, 120,c 12, 24, 36, 48, 60, 72, 84, 96, 108, 120,c The common multiples are in blue. The least common multiple is 60. EX A MPL E 2 3 Using the Least Common Multiple to Add Two Fractions Find: Solution r 8 5 + 15 12 We use the LCM of the denominators of the fractions and rewrite each fraction using the LCM as a common denominator. The LCM of the denominators (12 and 15) is 60. Rewrite each fraction using 60 as the denominator. 8 5 8 #4 5 #5 + = + 15 12 15 4 12 5 32 25 = + 60 60 32 + 25 = 60 57 = 60 19 = 20 Now Work PROBLEM 77 r SECTION R.1 Real Numbers 15 Historical Feature T he real number system has a history that stretches back at least to the ancient Babylonians (1800 BC). It is remarkable how much the ancient Babylonian attitudes resemble our own. As we stated in the text, the fundamental difficulty with irrational numbers is that they cannot be written as quotients of integers or, equivalently, as repeating or terminating decimals. The Babylonians wrote their numbers in a system based on 60 in the same way that we write ours based on 10. They would carry as many places for p as the accuracy of the problem demanded, just as we now use p ≈ 3 1 7 or p ≈ 3.1416 or p ≈ 3.14159 or p ≈ 3.14159265358979 depending on how accurate we need to be. Things were very different for the Greeks, whose number system allowed only rational numbers. When it was discovered that 12 was not a rational number, this was regarded as a fundamental flaw in the number concept. So serious was the matter that the Pythagorean Brotherhood (an early mathematical society) is said to have drowned one of its members for revealing this terrible secret. Greek mathematicians then turned away from the number concept, expressing facts about whole numbers in terms of line segments. In astronomy, however, Babylonian methods, including the Babylonian number system, continued to be used. Simon Stevin (1548–1620), probably using the Babylonian system as a model, invented the decimal system, complete with rules of calculation, in 1585. [Others, for example, al-Kashi of Samarkand (d. 1429), had made some progress in the same direction.] The decimal system so effectively conceals the difficulties that the need for more logical precision began to be felt only in the early 1800s. Around 1880, Georg Cantor (1845–1918) and Richard Dedekind (1831–1916) gave precise definitions of real numbers. Cantor’s definition, although more abstract and precise, has its roots in the decimal (and hence Babylonian) numerical system. Sets and set theory were a spin-off of the research that went into clarifying the foundations of the real number system. Set theory has developed into a large discipline of its own, and many mathematicians regard it as the foundation upon which modern mathematics is built. Cantor’s discoveries that infinite sets can also be counted and that there are different sizes of infinite sets are among the most astounding results of modern mathematics. R.1 Assess Your Understanding Concepts and Vocabulary 1. The numbers in the set e x ` x = and b ≠ 0 f are called a , where a, b are integers b numbers. 2. The value of the expression 4 + 5 # 6 - 3 is . 3. The fact that 2x + 3x = 12 + 32x is a consequence of the Property. 4. “The product of 5 and x + 3 equals 6” may be written as . 5. The intersection of sets A and B is denoted by which of the following? (a) A ∩ B (b) A ∪ B (c) A ⊆ B (d) A ∅ B 6. Choose the correct name for the set of numbers 5 0, 1, 2, 3, c6 . (a) Counting numbers (b) Whole numbers (c) Integers (d) Irrational numbers 7. True or False Rational numbers have decimals that either terminate or are nonterminating with a repeating block of digits. 8. True or False The Zero-Product Property states that the product of any number and zero equals zero. 9. True or False The least common multiple of 12 and 18 is 6. 10. True or False No real number is both rational and irrational. Skill Building In Problems 11–22, use U = universal set = 50, 1, 2, 3, 4, 5, 6, 7, 8, 96, A = 51, 3, 4, 5, 96 , B = 52, 4, 6, 7, 86 , and C = 51, 3, 4, 66 to find each set. 15. 1A ∪ B2 ∩ C 16. 1A ∩ B2 ∪ C 13. A ∩ B 14. A ∩ C 17. A 18. C 19. A ∩ B 20. B ∪ C 21. A ∪ B 22. B ∩ C 11. A ∪ B 12. A ∪ C In Problems 23–28, list the numbers in each set that are (a) Natural numbers, (b) Integers, (c) Rational numbers, (d) Irrational numbers, (e) Real numbers. 1 23. A = e - 6, , - 1.333c1the 3>s repeat2, p, 2, 5 f 2 5 24. B = e - , 2.060606c1the block 06 repeats2, 1.25, 0, 1, 25 f 3 1 1 1 25. C = e 0, 1, , , f 2 3 4 26. D = 5 - 1, - 1.1, - 1.2, - 1.36 27. E = e 22, p, 22 + 1, p + 1 f 2 28. F = e - 22, p + 22, 1 + 10.3 f 2 16 CHAPTER R Review In Problems 29–40, approximate each number (a) rounded and (b) truncated to three decimal places. 29. 18.9526 30. 25.86134 31. 28.65319 35. 9.9985 36. 1.0006 37. 3 7 32. 99.05249 38. 5 9 33. 0.06291 39. 34. 0.05388 521 15 40. 81 5 In Problems 41–50, write each statement using symbols. 41. The sum of 3 and 2 equals 5. 42. The product of 5 and 2 equals 10. 43. The sum of x and 2 is the product of 3 and 4. 44. The sum of 3 and y is the sum of 2 and 2. 45. The product of 3 and y is the sum of 1 and 2. 46. The product of 2 and x is the product of 4 and 6. 47. The difference x less 2 equals 6. 48. The difference 2 less y equals 6. 49. The quotient x divided by 2 is 6. 50. The quotient 2 divided by x is 6. In Problems 51–88, evaluate each expression. 54. 8 - 4 # 2 51. 9 - 4 + 2 52. 6 - 4 + 3 53. - 6 + 4 # 3 55. 4 + 5 - 8 56. 8 - 3 - 4 57. 4 + 59. 6 - 33 # 5 + 2 # 13 - 22 4 60. 2 # 38 - 314 + 22 4 - 3 61. 2 # 13 - 52 + 8 # 2 - 1 63. 10 - 36 - 2 # 2 + 18 - 32 4 # 2 65. 15 - 32 58. 2 - 1 2 62. 1 - 14 # 3 - 2 + 22 64. 2 - 5 # 4 - 36 # 13 - 42 4 66. 15 + 42 1 2 1 3 1 3 67. 4 + 8 5 - 3 68. 2 - 4 5 - 3 69. 3 # 10 5 21 70. 5# 3 9 10 71. 6 # 10 25 27 72. 21 # 100 25 3 73. 3 2 + 4 5 74. 4 1 + 3 2 75. 5 9 + 6 5 76. 8 15 + 9 2 77. 5 1 + 18 12 78. 2 8 + 15 9 79. 1 7 30 18 80. 3 2 14 21 3 2 81. 20 15 85. 7 1#3 + 2 5 10 6 3 82. 35 14 5 18 83. 11 27 2 4 1 + # 3 5 6 87. 2 # 86. 5 21 84. 2 35 3 3 + 4 8 88. 3 # 5 1 6 2 In Problems 89–100, use the Distributive Property to remove the parentheses. 89. 61x + 42 90. 412x - 12 91. x1x - 42 92. 4x1x + 32 1 3 93. 2a x - b 4 2 2 1 94. 3a x + b 3 6 95. 1x + 22 1x + 42 96. 1x + 52 1x + 12 97. 1x - 22 1x + 12 98. 1x - 42 1x + 12 99. 1x - 82 1x - 22 100. 1x - 42 1x - 22 Explaining Concepts: Discussion and Writing 101. Explain to a friend how the Distributive Property is used to justify the fact that 2x + 3x = 5x. 102. Explain to a friend 12 + 32 # 4 = 20. why 2 + 3 # 4 = 14, whereas 103. Explain why 213 # 42 is not equal to 12 # 32 # 12 # 42. 104. Explain why 4 + 3 4 3 is not equal to + . 2 + 5 2 5 SECTION R.2 Algebra Essentials 105. Is subtraction commutative? Support your conclusion with an example. 106. Is subtraction associative? Support your conclusion with an example. 107. Is division commutative? Support your conclusion with an example. 108. Is division associative? Support your conclusion with an example. 109. If 2 = x, why does x = 2? 17 113. A rational number is defined as the quotient of two integers. When written as a decimal, the decimal will either repeat or terminate. By looking at the denominator of the rational number, there is a way to tell in advance whether its decimal representation will repeat or terminate. Make a list of rational numbers and their decimals. See if you can discover the pattern. Confirm your conclusion by consulting books on number theory at the library. Write a brief essay on your findings. 114. The current time is 12 noon CST. What time (CST) will it be 12,997 hours from now? 110. If x = 5, why does x2 + x = 30? 111. Are there any real numbers that are both rational and irrational? Are there any real numbers that are neither? Explain your reasoning. 0 a 1a ≠ 02 and are undefined, but for different 0 0 reasons. Write a paragraph or two explaining the different reasons. 115. Both 112. Explain why the sum of a rational number and an irrational number must be irrational. R.2 Algebra Essentials OBJECTIVES 1 Graph Inequalities (p. 18) 2 Find Distance on the Real Number Line (p. 19) 3 Evaluate Algebraic Expressions (p. 20) 4 Determine the Domain of a Variable (p. 21) 5 Use the Laws of Exponents (p. 21) 6 Evaluate Square Roots (p. 23) 7 Use a Calculator to Evaluate Exponents (p. 24) 8 Use Scientific Notation (p. 24) The Real Number Line 2 units Scale 1 unit O 3 2 1 1–2 0 –12 1 2 2 3 Figure 8 Real number line DEFINITION Real numbers can be represented by points on a line called the real number line. There is a one-to-one correspondence between real numbers and points on a line. That is, every real number corresponds to a point on the line, and each point on the line has a unique real number associated with it. Pick a point on a line somewhere in the center, and label it O. This point, called the origin, corresponds to the real number 0. See Figure 8. The point 1 unit to the right of O corresponds to the number 1. The distance between 0 and 1 determines the scale of the number line. For example, the point associated with the number 2 is twice as far from O as 1. Notice that an arrowhead on the right end of the line indicates the direction in which the numbers increase. Points to the left of the origin correspond to the real numbers - 1, - 2, and so on. Figure 8 also shows 1 1 the points associated with the rational numbers - and and with the irrational 2 2 numbers 12 and p. The real number associated with a point P is called the coordinate of P, and the line whose points have been assigned coordinates is called the real number line. Now Work PROBLEM 13 18 CHAPTER R Review O 3 2 3–2 1 1–2 Negative real numbers 0 1 – 2 Zero 1 3– 2 2 3 Positive real numbers Figure 9 The real number line consists of three classes of real numbers, as shown in Figure 9. 1. The negative real numbers are the coordinates of points to the left of the origin O. 2. The real number zero is the coordinate of the origin O. 3. The positive real numbers are the coordinates of points to the right of the origin O. Multiplication Properties of Positive and Negative Numbers 1. The product of two positive numbers is a positive number. 2. The product of two negative numbers is a positive number. 3. The product of a positive number and a negative number is a negative number. 1 Graph Inequalities a (a) a b b a b (b) a b b (c) a b a Figure 10 EX AMPLE 1 An important property of the real number line follows from the fact that, given two numbers (points) a and b, either a is to the left of b, or a is at the same location as b, or a is to the right of b. See Figure 10. If a is to the left of b, then “a is less than b,” which is written a 6 b. If a is to the right of b, then “a is greater than b,” which is written a 7 b. If a is at the same location as b, then a = b. If a is either less than or equal to b, then a … b. Similarly, a Ú b means that a is either greater than or equal to b. Collectively, the symbols 6 , 7 , … , and Ú are called inequality symbols. Note that a 6 b and b 7 a mean the same thing. It does not matter whether we write 2 6 3 or 3 7 2. Furthermore, if a 6 b or if b 7 a, then the difference b - a is positive. Do you see why? Using Inequality Symbols (a) 3 6 7 (d) - 8 6 - 4 (b) - 8 7 - 16 (e) 4 7 - 1 (c) - 6 6 0 (f) 8 7 0 r In Example 1(a), we conclude that 3 6 7 either because 3 is to the left of 7 on the real number line or because the difference, 7 - 3 = 4, is a positive real number. Similarly, we conclude in Example 1(b) that - 8 7 - 16 either because - 8 lies to the right of - 16 on the real number line or because the difference, - 8 - 1 - 162 = - 8 + 16 = 8, is a positive real number. Look again at Example 1. Note that the inequality symbol always points in the direction of the smaller number. An inequality is a statement in which two expressions are related by an inequality symbol. The expressions are referred to as the sides of the inequality. Inequalities of the form a 6 b or b 7 a are called strict inequalities, whereas inequalities of the form a … b or b Ú a are called nonstrict inequalities. Based on the discussion so far, we conclude that a 7 0 is equivalent to a is positive a 6 0 is equivalent to a is negative SECTION R.2 Algebra Essentials 19 We sometimes read a 7 0 by saying that “a is positive.” If a Ú 0, then either a 7 0 or a = 0, and we may read this as “a is nonnegative.” Now Work PROBLEMS 17 AND 27 Graphing Inequalities EXAMPL E 2 (a) On the real number line, graph all numbers x for which x 7 4. (b) On the real number line, graph all numbers x for which x … 5. Solution –2 –1 0 1 2 3 4 5 6 7 3 4 5 6 7 (a) See Figure 11. Notice that we use a left parenthesis to indicate that the number 4 is not part of the graph. (b) See Figure 12. Notice that we use a right bracket to indicate that the number 5 is part of the graph. r Figure 11 x 7 4 Now Work 2 1 0 1 2 PROBLEM 33 Figure 12 x … 5 2 Find Distance on the Real Number Line 4 units 3 units 5 4 3 2 1 0 1 2 3 4 The absolute value of a number a is the distance from 0 to a on the number line. For example, - 4 is 4 units from 0, and 3 is 3 units from 0. See Figure 13. That is, the absolute value of - 4 is 4, and the absolute value of 3 is 3. A more formal definition of absolute value is given next. Figure 13 DEFINITION The absolute value of a real number a, denoted by the symbol 0 a 0 , is defined by the rules 0 a 0 = a if a Ú 0 and 0 a 0 = - a if a 6 0 For example, because - 4 6 0, the second rule must be used to get 0 - 4 0 = - 1 - 42 = 4. EXAMPL E 3 Computing Absolute Value (a) 0 8 0 = 8 (b) 0 0 0 = 0 (c) 0 - 15 0 = - 1 - 152 = 15 r Look again at Figure 13. The distance from - 4 to 3 is 7 units. This distance is the difference 3 - 1 - 42, obtained by subtracting the smaller coordinate from the larger. However, since 0 3 - 1 - 42 0 = 0 7 0 = 7 and 0 - 4 - 3 0 = 0 - 7 0 = 7, we can use absolute value to calculate the distance between two points without being concerned about which is smaller. DEFINITION If P and Q are two points on a real number line with coordinates a and b, respectively, the distance between P and Q, denoted by d 1P, Q2, is d 1P, Q2 = 0 b - a 0 Since 0 b - a 0 = 0 a - b 0 , it follows that d 1P, Q2 = d 1Q, P2. 20 CHAPTER R Review EX AMPLE 4 Finding Distance on a Number Line Let P, Q, and R be points on a real number line with coordinates - 5, 7, and - 3, respectively. Find the distance (a) between P and Q Solution (b) between Q and R See Figure 14. P R Q 5 4 3 2 1 0 1 2 3 4 5 6 7 d (P, Q) ⏐7 (5)⏐ 12 d(Q, R) ⏐ 3 7 ⏐ 10 Figure 14 (a) d 1P, Q2 = 0 7 - 1 - 52 0 = 0 12 0 = 12 (b) d 1Q, R2 = 0 - 3 - 7 0 = 0 - 10 0 = 10 Now Work PROBLEM r 39 3 Evaluate Algebraic Expressions Remember, in algebra we use letters such as x, y, a, b, and c to represent numbers. If the letter used is to represent any number from a given set of numbers, it is called a variable. A constant is either a fixed number, such as 5 or 13, or a letter that represents a fixed (possibly unspecified) number. Constants and variables are combined using the operations of addition, subtraction, multiplication, and division to form algebraic expressions. Examples of algebraic expressions include 3 7x - 2y 1 - t To evaluate an algebraic expression, substitute a numerical value for each variable. x + 3 EX AMPLE 5 Evaluating an Algebraic Expression Evaluate each expression if x = 3 and y = - 1. (a) x + 3y Solution (b) 5xy (c) (d) 0 - 4x + y 0 3y 2 - 2x (a) Substitute 3 for x and - 1 for y in the expression x + 3y. x + 3y = 3 + 31 - 12 = 3 + 1 - 32 = 0 c x = 3, y = - 1 (b) If x = 3 and y = - 1, then 5xy = 5132 1 - 12 = - 15 (c) If x = 3 and y = - 1, then 31 - 12 3y -3 -3 3 = = = = 2 - 2x 2 - 2132 2 - 6 -4 4 (d) If x = 3 and y = - 1, then 0 - 4x + y 0 = 0 - 4132 + 1 - 12 0 = 0 - 12 + 1 - 12 0 = 0 - 13 0 = 13 Now Work PROBLEMS 41 AND 49 r SECTION R.2 Algebra Essentials 21 4 Determine the Domain of a Variable In working with expressions or formulas involving variables, the variables may be allowed to take on values from only a certain set of numbers. For example, in the formula for the area A of a circle of radius r, A = pr 2, the variable r is necessarily 1 restricted to the positive real numbers. In the expression , the variable x cannot x take on the value 0, since division by 0 is not defined. DEFINITION EXAMPL E 6 The set of values that a variable may assume is called the domain of the variable. Finding the Domain of a Variable The domain of the variable x in the expression 5 x - 2 is 5 x 0 x ≠ 26 since, if x = 2, the denominator becomes 0, which is not defined. EXAMPL E 7 r Circumference of a Circle In the formula for the circumference C of a circle of radius r, C = 2pr the domain of the variable r, representing the radius of the circle, is the set of positive real numbers, 5 r 0 r 7 06 . The domain of the variable C, representing the circumference of the circle, is also the set of positive real numbers, 5 C 0 C 7 06 . r In describing the domain of a variable, we may use either set notation or words, whichever is more convenient. Now Work PROBLEM 59 5 Use the Laws of Exponents Integer exponents provide a shorthand notation for representing repeated multiplications of a real number. For example, 34 = 3 # 3 # 3 # 3 = 81 Additionally, many formulas have exponents. For example, r The formula for the horsepower rating H of an engine is H = D2 N 2.5 where D is the diameter of a cylinder and N is the number of cylinders. r A formula for the resistance R of blood flowing in a blood vessel is R = C L r4 where L is the length of the blood vessel, r is the radius, and C is a positive constant. 22 CHAPTER R Review DEFINITION If a is a real number and n is a positive integer, then the symbol an represents the product of n factors of a. That is, an = a # a # c # a u (1) n factors In the definition it is understood that a1 = a. Furthermore, a2 = a # a, a3 = a # a # a, and so on. In the expression an, a is called the base and n is called the exponent, or power. We read an as “a raised to the power n” or as “a to the nth power.” We usually - 24 = - 1 # 24 = - 16 read a2 as “a squared” and a3 as “a cubed.” whereas In working with exponents, the operation of raising to a power is performed 1- 22 4 = 1- 221- 22 1- 22 1- 22 = 16 before any other operation. As examples, WARNING Be careful with minus signs and exponents. ■ 4 # 32 = 4 # 9 = 36 22 + 32 = 4 + 9 = 13 - 24 = - 16 5 # 32 + 2 # 4 = 5 # 9 + 2 # 4 = 45 + 8 = 53 Parentheses are used to indicate operations to be performed first. For example, 1 - 22 4 = 1 - 22 1 - 22 1 - 22 1 - 22 = 16 DEFINITION 12 + 32 2 = 52 = 25 If a ≠ 0, then a0 = 1 DEFINITION If a ≠ 0 and if n is a positive integer, then a-n = 1 an Whenever you encounter a negative exponent, think “reciprocal.” EX AMPLE 8 Evaluating Expressions Containing Negative Exponents (a) 2-3 = 1 1 = 8 23 Now Work (b) x -4 = PROBLEMS 1 -2 (c) a b = 5 1 x4 77 AND 1 1 2 a b 5 = 1 = 25 1 25 r 97 The following properties, called the Laws of Exponents, can be proved using the preceding definitions. In the list, a and b are real numbers, and m and n are integers. THEOREM Laws of Exponents am an = am + n 1am 2 = amn 1 am m-n = n-m n = a a a n if a ≠ 0 1ab2 n = an bn a n an a b = n b b if b ≠ 0 SECTION R.2 Algebra Essentials EXAMPL E 9 Using the Laws of Exponents (a) x -3 # x5 = x -3 + 5 = x2 x ≠ 0 (b) 1x -3 2 = x -3 2 = x -6 = (c) 12x2 3 = 23 # x3 = 8x3 2 4 24 16 (d) a b = 4 = 3 81 3 (e) 2 # 1 x6 x ≠ 0 x -2 = x -2 - 1-52 = x3 x ≠ 0 x -5 Now Work EX AM PL E 1 0 23 PROBLEM r 79 Using the Laws of Exponents Write each expression so that all exponents are positive. (a) Solution (a) x5 y -2 x3 y x5 y -2 x3 y (b) ¢ x ≠ 0, y ≠ 0 = (b) ¢ x -3 -2 ≤ x ≠ 0, y ≠ 0 3y -1 x2 x5 # y -2 5 - 3 # -2 - 1 2 -3 2# 1 y = x y = x = = x x3 y y3 y3 1x -3 2 -2 x -3 -2 x6 x6 9x6 ≤ = = = = 1 2 y2 3y -1 13y -1 2 -2 3-2 1y -1 2 -2 y 9 Now Work PROBLEM r 89 6 Evaluate Square Roots In Words 136 means “give me the nonnegative number whose square is 36.” DEFINITION A real number is squared when it is raised to the power 2. The inverse of squaring is finding a square root. For example, since 62 = 36 and 1 - 62 2 = 36, the numbers 6 and - 6 are square roots of 36. The symbol 1 , called a radical sign, is used to denote the principal, or nonnegative, square root. For example, 136 = 6. If a is a nonnegative real number, the nonnegative number b such that b2 = a is the principal square root of a, and is denoted by b = 1a. The following comments are noteworthy: 1. Negative numbers do not have square roots (in the real number system), because the square of any real number is nonnegative. For example, 1- 4 is not a real number, because there is no real number whose square is - 4. 2. The principal square root of 0 is 0, since 02 = 0. That is, 10 = 0. 3. The principal square root of a positive number is positive. 4. If c Ú 0, then 1 1c2 2 = c. For example, 1 122 2 = 2 and 1 132 2 = 3. EX AM PL E 11 Evaluating Square Roots (a) 264 = 8 (b) 1 1 = 4 A 16 (c) 1 21.42 2 = 1.4 r Examples 11(a) and (b) are examples of square roots of perfect squares, since 1 1 2 64 = 82 and = a b . 16 4 24 CHAPTER R Review Consider the expression 2a2. Since a2 Ú 0, the principal square root of a2 is defined whether a 7 0 or a 6 0. However, since the principal square root is nonnegative, we need an absolute value to ensure the nonnegative result. That is, 2a2 = 0 a 0 EX A MPL E 1 2 a any real number Using Equation (2) (a) 2 12.32 2 = 0 2.3 0 = 2.3 (2) (b) 2 1 - 2.32 2 = 0 - 2.3 0 = 2.3 (c) 2x2 = 0 x 0 r Now Work PROBLEM 85 7 Use a Calculator to Evaluate Exponents Your calculator has either a caret key, ^ , or an xy key, that is used for computations involving exponents. EX A MPL E 1 3 Exponents on a Graphing Calculator Evaluate: 12.32 5 Solution Figure 15 shows the result using a TI-84 Plus C graphing calculator. Now Work PROBLEM r 115 8 Use Scientific Notation Measurements of physical quantities can range from very small to very large. For example, the mass of a proton is approximately 0.00000000000000000000000000167 kilogram and the mass of Earth is about 5,980,000,000,000,000,000,000,000 kilograms. These numbers obviously are tedious to write down and difficult to read, so we use exponents to rewrite them. Figure 15 DEFINITION When a number has been written as the product of a number x, where 1 … x 6 10, times a power of 10, it is said to be written in scientific notation. In scientific notation, Mass of a proton = 1.67 * 10-27 kilogram Mass of Earth = 5.98 * 1024 kilograms Converting a Decimal to Scientific Notation To change a positive number into scientific notation: 1. Count the number N of places that the decimal point must be moved to arrive at a number x, where 1 … x 6 10. 2. If the original number is greater than or equal to 1, the scientific notation is x * 10N. If the original number is between 0 and 1, the scientific notation is x * 10-N. SECTION R.2 Algebra Essentials EX AM PL E 1 4 25 Using Scientific Notation Write each number in scientific notation. (a) 9582 Solution (b) 1.245 (c) 0.285 (d) 0.000561 (a) The decimal point in 9582 follows the 2. Count left from the decimal point 9 5 c 8 c 2 # c 3 2 1 stopping after three moves, because 9.582 is a number between 1 and 10. Since 9582 is greater than 1, we write 9582 = 9.582 * 103 (b) The decimal point in 1.245 is between the 1 and 2. Since the number is already between 1 and 10, the scientific notation for it is 1.245 * 100 = 1.245. (c) The decimal point in 0.285 is between the 0 and the 2. We count 0 2 # 8 5 c 1 stopping after one move, because 2.85 is a number between 1 and 10. Since 0.285 is between 0 and 1, we write 0.285 = 2.85 * 10-1 (d) The decimal point in 0.000561 is moved as follows: 0 0 # 0 c 1 0 c 2 5 c 3 6 1 c 4 As a result, r 0.000561 = 5.61 * 10-4 Now Work EX AM PL E 15 121 PROBLEM Changing from Scientific Notation to Decimals Write each number as a decimal. (a) 2.1 * 104 Solution (b) 3.26 * 10-5 (a) 2.1 * 104 = 2 1 # 0 c (b) 3.26 * 10-5 = 0 (c) 1 * 10 = 0 0 c c 1 2 3 4 0 0 0 c 5 1 c 4 * 104 = 21,000 0 c 0 c -2 0 c c 3 # Now Work # 2 6 * 10-5 = 0.0000326 c 1 0 * 10 -2 = 0.01 r 1 PROBLEM 3 2 c 2 (c) 1 * 10-2 129 26 CHAPTER R Review EX A MPL E 1 6 Using Scientific Notation (a) The diameter of the smallest living cell is only about 0.00001 centimeter (cm).* Express this number in scientific notation. (b) The surface area of Earth is about 1.97 * 108 square miles.† Express the surface area as a whole number. Solution (a) 0.00001 cm = 1 * 10-5 cm because the decimal point is moved five places and the number is less than 1. (b) 1.97 * 108 square miles = 197,000,000 square miles. r Now Work PROBLEM 155 COMMENT On a calculator, a number such as 3.615 * 1012 is usually displayed as 3.615E12. ■ * Powers of Ten, Philip and Phylis Morrison. † 2011 Information Please Almanac. Historical Feature T he word algebra is derived from the Arabic word al-jabr. This word is a part of the title of a ninth-century work, “Hisâb al-jabr w’al-muqâbalah,” written by Mohammed ibn Músâ al-Khwârizmî. The word al-jabr means “a restoration,” a reference to the fact that if a number is added to one side of an equation, then it must also be added to the other side in order to “restore” the equality. The title of the work, freely translated, is“The Science of Reduction and Cancellation.”Of course, today, algebra has come to mean a great deal more. R.2 Assess Your Understanding Concepts and Vocabulary 1. A(n) is a letter used in algebra to represent any number from a given set of numbers. 2. On the real number line, the real number zero is the coordinate of the . 3. An inequality of the form a 7 b is called a(n) inequality. 4. In the expression 24, the number 2 is called the is called the . 5. In scientific notation, 1234.5678 = and 4 . 6. If a is a nonnegative real number, then which inequality statement best describes a? (a) a 6 0 (b) a 7 0 (c) a … 0 (d) a Ú 0 7. Let a and b be non-zero real numbers and m and n be integers. Which of the following is not a law of exponents? an a n (a) a b = n b b am (c) n = am - n a (b) 1am 2 n = am + n (d) 1ab2 n = anbn 8. True or False The product of two negative real numbers is always greater than zero. 9. True or False The distance between two distinct points on the real number line is always greater than zero. 10. True or False The absolute value of a real number is always greater than zero. 11. True or False When a number is expressed in scientific notation, it is expressed as the product of a number x, 0 … x 6 1, and a power of 10. 12. True or False To multiply two expressions having the same base, retain the base and multiply the exponents. Skill Building 5 3 13. On the real number line, label the points with coordinates 0, 1, - 1, , - 2.5, , and 0.25. 2 4 3 1 2 14. Repeat Problem 13 for the coordinates 0, - 2, 2, - 1.5, , , and . 2 3 3 In Problems 15–24, replace the question mark by 6, 7, or =, whichever is correct. 15. 1 ?0 2 20. 22 ? 1.41 16. 5 ? 6 21. 1 ? 0.5 2 17. - 1 ? - 2 22. 1 ? 0.33 3 18. - 3 ? 23. 2 ? 0.67 3 5 2 19. p ? 3.14 24. 1 ? 0.25 4 SECTION R.2 Algebra Essentials 27 In Problems 25–30, write each statement as an inequality. 25. x is positive 26. z is negative 27. x is less than 2 28. y is greater than - 5 29. x is less than or equal to 1 30. x is greater than or equal to 2 In Problems 31–34, graph the numbers x on the real number line. 32. x 6 4 31. x Ú - 2 34. x … 7 33. x 7 - 1 In Problems 35–40, use the given real number line to compute each distance. A 35. d1C, D2 B C D 4 3 2 1 0 1 36. d1C, A2 E 2 37. d1D, E2 3 4 5 6 38. d1C, E2 39. d1A, E2 40. d1D, B2 In Problems 41–48, evaluate each expression if x = - 2 and y = 3. 41. x + 2y 45. 42. 3x + y 2x x - y 46. x + y x - y 47. 3x + 2y 2 + y In Problems 49–58, find the value of each expression if x = 3 and y = - 2. 49. 0 x + y 0 54. 0y0 y 44. - 2x + xy 43. 5xy + 2 48. 2x - 3 y 0x0 50. 0 x - y 0 51. 0 x 0 + 0 y 0 52. 0 x 0 - 0 y 0 53. 55. 0 4x - 5y 0 56. 0 3x + 2y 0 57. 0 0 4x 0 - 0 5y 0 0 58. 3 0 x 0 + 2 0 y 0 x In Problems 59–66, determine which of the values (a) through (d), if any, must be excluded from the domain of the variable in each expression. (a) x = 3 (b) x = 1 (c) x = 0 (d) x = - 1 59. x2 - 1 x 60. x2 + 1 x 61. x x2 - 9 62. x x2 + 9 63. x2 x + 1 64. x3 x - 1 65. x2 + 5x - 10 x3 - x 66. - 9x2 - x + 1 x3 + x 70. x - 2 x - 6 2 2 In Problems 67–70, determine the domain of the variable x in each expression. x 4 -6 69. 67. 68. x + 4 x - 5 x + 4 5 In Problems 71–74, use the formula C = 1F - 322 for converting degrees Fahrenheit into degrees Celsius to find the Celsius measure 9 of each Fahrenheit temperature. 71. F = 32° 72. F = 212° 73. F = 77° 74. F = - 4° In Problems 75–86, simplify each expression. 75. 1 - 42 2 76. - 42 77. 4-2 78. - 4-2 79. 3-6 # 34 80. 4-2 # 43 81. 13-2 2 -1 82. 12-1 2 -3 83. 225 84. 236 85. 21 - 42 2 86. 21 - 32 2 In Problems 87–96, simplify each expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative, we assume that the base is not 0. 87. 18x3 2 92. x -2 y xy2 2 88. 1 - 4x2 2 - 1 93. 1 - 22 3 x4 1yz2 2 32 xy3 z 89. 1x2 y -1 2 94. 2 4x -2 1yz2 -1 23 x4 y 90. 1x -1 y2 95. ¢ 3x -1 4y 3 ≤ -1 -2 91. x2 y 3 xy4 96. ¢ 5x -2 -3 ≤ 6y -2 28 CHAPTER R Review In Problems 97–108, find the value of each expression if x = 2 and y = - 1. 99. x2 + y2 100. x2 y2 97. 2xy -1 98. - 3x -1 y 101. 1xy2 2 102. 1x + y2 2 103. 2x2 104. 1 1x2 2 105. 2x2 + y2 106. 2x2 + 2y2 107. xy 108. yx 109. Find the value of the expression 2x3 - 3x2 + 5x - 4 if x = 2. What is the value if x = 1? 110. Find the value of the expression 4x3 + 3x2 - x + 2 if x = 1. What is the value if x = 2? 111. What is the value of 16662 4 12222 4 112. What is the value of 10.12 3 1202 3 ? ? In Problems 113–120, use a calculator to evaluate each expression. Round your answer to three decimal places. 113. 18.22 6 114. 13.72 5 115. 16.12 -3 116. 12.22 -5 117. 1 - 2.82 6 118. - 12.82 6 119. 1 - 8.112 -4 120. - 18.112 -4 In Problems 121–128, write each number in scientific notation. 121. 454.2 122. 32.14 123. 0.013 124. 0.00421 125. 32,155 126. 21,210 127. 0.000423 128. 0.0514 In Problems 129–136, write each number as a decimal. 129. 6.15 * 104 130. 9.7 * 103 131. 1.214 * 10-3 132. 9.88 * 10-4 133. 1.1 * 108 134. 4.112 * 102 135. 8.1 * 10-2 136. 6.453 * 10-1 Applications and Extensions In Problems 137–146, express each statement as an equation involving the indicated variables. 137. Area of a Rectangle The area A of a rectangle is the product of its length l and its width w. l A 141. Area of an Equilateral Triangle The area A of an 13 equilateral triangle is times the square of the length x of 4 one side. w x x 138. Perimeter of a Rectangle The perimeter P of a rectangle is twice the sum of its length l and its width w. 139. Circumference of a Circle The circumference C of a circle is the product of p and its diameter d. C x 142. Perimeter of an Equilateral Triangle The perimeter P of an equilateral triangle is 3 times the length x of one side. 4 143. Volume of a Sphere The volume V of a sphere is times p 3 times the cube of the radius r. d r 140. Area of a Triangle The area A of a triangle is one-half the product of its base b and its height h. h b 144. Surface Area of a Sphere The surface area S of a sphere is 4 times p times the square of the radius r. SECTION R.2 Algebra Essentials 145. Volume of a Cube The volume V of a cube is the cube of the length x of a side. of its products is a ball bearing with a stated radius of 3 centimeters (cm). Only ball bearings with a radius within 0.01 cm of this stated radius are acceptable. If x is the radius of a ball bearing, a formula describing this situation is 0 x - 3 0 … 0.01 x x x 146. Surface Area of a Cube The surface area S of a cube is 6 times the square of the length x of a side. 147. Manufacturing Cost The weekly production cost C of manufacturing x watches is given by the formula C = 4000 + 2x, where the variable C is in dollars. (a) What is the cost of producing 1000 watches? (b) What is the cost of producing 2000 watches? 148. Balancing a Checkbook At the beginning of the month, Mike had a balance of $210 in his checking account. During the next month, he deposited $80, wrote a check for $120, made another deposit of $25, and wrote two checks: one for $60 and the other for $32. He was also assessed a monthly service charge of $5. What was his balance at the end of the month? 29 (a) Is a ball bearing of radius x = 2.999 acceptable? (b) Is a ball bearing of radius x = 2.89 acceptable? 154. Body Temperature Normal human body temperature is 98.6°F. A temperature x that differs from normal by at least 1.5°F is considered unhealthy. A formula that describes this is 0 x - 98.6 0 Ú 1.5 (a) Show that a temperature of 97°F is unhealthy. (b) Show that a temperature of 100°F is not unhealthy. 155. Distance from Earth to Its Moon The distance from Earth to the Moon is about 4 * 108 meters.* Express this distance as a whole number. 156. Height of Mt. Everest The height of Mt. Everest is 8848 meters.* Express this height in scientific notation. 157. Wavelength of Visible Light The wavelength of visible light is about 5 * 10-7 meter.* Express this wavelength as a decimal. In Problems 149 and 150, write an inequality using an absolute value to describe each statement. 158. Diameter of an Atom The diameter of an atom is about 1 * 10-10 meter.* Express this diameter as a decimal. 149. x is at least 6 units from 4. 159. Diameter of Copper Wire The smallest commercial copper wire is about 0.0005 inch in diameter.† Express this diameter using scientific notation. 150. x is more than 5 units from 2. 151. U.S. Voltage In the United States, normal household voltage is 110 volts. It is acceptable for the actual voltage x to differ from normal by at most 5 volts. A formula that describes this is 0 x - 110 0 … 5 (a) Show that a voltage of 108 volts is acceptable. (b) Show that a voltage of 104 volts is not acceptable. 152. Foreign Voltage In other countries, normal household voltage is 220 volts. It is acceptable for the actual voltage x to differ from normal by at most 8 volts. A formula that describes this is 0 x - 220 0 … 8 (a) Show that a voltage of 214 volts is acceptable. (b) Show that a voltage of 209 volts is not acceptable. 153. Making Precision Ball Bearings The FireBall Company manufactures ball bearings for precision equipment. One 160. Smallest Motor The smallest motor ever made is less than 0.05 centimeter wide.† Express this width using scientific notation. 161. Astronomy One light-year is defined by astronomers to be the distance that a beam of light will travel in 1 year (365 days). If the speed of light is 186,000 miles per second, how many miles are in a light-year? Express your answer in scientific notation. 162. Astronomy How long does it take a beam of light to reach Earth from the Sun when the Sun is 93,000,000 miles from Earth? Express your answer in seconds, using scientific notation. 163. Does 1 equal 0.333? If not, which is larger? By how much? 3 164. Does 2 equal 0.666? If not, which is larger? By how much? 3 Explaining Concepts: Discussion and Writing 165. Is there a positive real number “closest” to 0? 166. Number game I’m thinking of a number! It lies between 1 and 10; its square is rational and lies between 1 and 10. The number is larger than p. Correct to two decimal places (that is, truncated to two decimal places), name the number. Now think of your own number, describe it, and challenge a fellow student to name it. * † Powers of Ten, Philip and Phylis Morrison. 2011 Information Please Almanac. 167. Write a brief paragraph that illustrates the similarities and differences between “less than” 1 6 2 and “less than or equal to” 1 … 2. 168. Give a reason why the statement 5 6 8 is true. 30 CHAPTER R Review R.3 Geometry Essentials OBJECTIVES 1 Use the Pythagorean Theorem and Its Converse (p. 30) 2 Know Geometry Formulas (p. 31) 3 Understand Congruent Triangles and Similar Triangles (p. 32) 1 Use the Pythagorean Theorem and Its Converse Hypotenuse c b Leg 90° a Leg The Pythagorean Theorem is a statement about right triangles. A right triangle is one that contains a right angle—that is, an angle of 90°. The side of the triangle opposite the 90° angle is called the hypotenuse; the remaining two sides are called legs. In Figure 16 we have used c to represent the length of the hypotenuse and a and b to represent the lengths of the legs. Notice the use of the symbol to show the 90° angle. We now state the Pythagorean Theorem. Figure 16 A right triangle PYTHAGOREAN THEOREM In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is, in the right triangle shown in Figure 16, c 2 = a 2 + b2 (1) A proof of the Pythagorean Theorem is given at the end of this section. EX AMPLE 1 Finding the Hypotenuse of a Right Triangle In a right triangle, one leg has length 4 and the other has length 3. What is the length of the hypotenuse? Solution Since the triangle is a right triangle, we use the Pythagorean Theorem with a = 4 and b = 3 to find the length c of the hypotenuse. From equation (1), c 2 = a 2 + b2 c 2 = 42 + 32 = 16 + 9 = 25 r c = 225 = 5 Now Work PROBLEM 15 The converse of the Pythagorean Theorem is also true. CONVERSE OF THE PYTHAGOREAN THEOREM In a triangle, if the square of the length of one side equals the sum of the squares of the lengths of the other two sides, the triangle is a right triangle. The 90° angle is opposite the longest side. A proof is given at the end of this section. EX A MPLE 2 Verifying That a Triangle Is a Right Triangle Show that a triangle whose sides are of lengths 5, 12, and 13 is a right triangle. Identify the hypotenuse. Solution Square the lengths of the sides. 52 = 25 122 = 144 132 = 169 SECTION R.3 Geometry Essentials 31 Notice that the sum of the first two squares (25 and 144) equals the third square (169). That is, because 52 + 122 = 132, the triangle is a right triangle. The longest side, 13, is the hypotenuse. See Figure 17. 13 r 5 90° Now Work PROBLEM 12 23 Figure 17 Applying the Pythagorean Theorem EXAMPL E 3 The tallest building in the world is Burj Khalifa in Dubai, United Arab Emirates, at 2717 feet and 163 floors. The observation deck is 1483 feet above ground level. How far can a person standing on the observation deck see (with the aid of a telescope)? Use 3960 miles for the radius of Earth. Source: Council on Tall Buildings and Urban Habitat Solution From the center of Earth, draw two radii: one through Burj Khalifa and the other to the farthest point a person can see from the observation deck. See Figure 18. Apply the Pythagorean Theorem to the right triangle. 1483 mile. Then Since 1 mile = 5280 feet, 1483 feet = 5280 d 2 + 139602 2 = a3960 + d 2 = a3960 + 1483 2 b 5280 1483 2 b - 139602 2 ≈ 2224.58 5280 d ≈ 47.17 A person can see more than 47 miles from the observation deck. d 1483 ft 3960 mi r Figure 18 Now Work PROBLEM 55 2 Know Geometry Formulas Certain formulas from geometry are useful in solving algebra problems. For a rectangle of length l and width w, w Area = lw Perimeter = 2l + 2w l For a triangle with base b and altitude h, h b Area = 1 bh 2 32 CHAPTER R Review For a circle of radius r (diameter d = 2r), d r Area = pr 2 Circumference = 2pr = pd For a closed rectangular box of length l, width w, and height h, h w l Volume = lwh Surface area = 2lh + 2wh + 2lw For a sphere of radius r, r Volume = 4 3 pr 3 Surface area = 4pr 2 For a closed right circular cylinder of height h and radius r, r h Volume = pr 2 h Now Work EX AMPLE 4 PROBLEM Surface area = 2pr 2 + 2prh 31 Using Geometry Formulas A Christmas tree ornament is in the shape of a semicircle on top of a triangle. How many square centimeters (cm2) of copper is required to make the ornament if the height of the triangle is 6 cm and the base is 4 cm? Solution 4 l6 See Figure 19. The amount of copper required equals the shaded area. This area is the sum of the areas of the triangle and the semicircle. The triangle has height h = 6 and base b = 4. The semicircle has diameter d = 4, so its radius is r = 2. Area = Area of triangle + Area of semicircle = Figure 19 1 1 1 1 bh + pr 2 = 142 162 + p # 22 2 2 2 2 b = 4; h = 6; r = 2 = 12 + 2p ≈ 18.28 cm2 About 18.28 cm2 of copper is required. Now Work PROBLEM r 49 3 Understand Congruent Triangles and Similar Triangles In Words Two triangles are congruent if they have the same size and shape. DEFINITION Throughout the text we will make reference to triangles. We begin with a discussion of congruent triangles. According to dictionary.com, the word congruent means “coinciding exactly when superimposed.” For example, two angles are congruent if they have the same measure, and two line segments are congruent if they have the same length. Two triangles are congruent if each pair of corresponding angles have the same measure and each pair of corresponding sides are the same length. In Figure 20, corresponding angles are equal and the corresponding sides are equal in length: a = d, b = e, and c = f . As a result, these triangles are congruent. SECTION R.3 Geometry Essentials 100 a 30 100 d b e 50 30 50 c Figure 20 33 f Congruent triangles Actually, it is not necessary to verify that all three angles and all three sides are the same measure to determine whether two triangles are congruent. Determining Congruent Triangles 1. Angle–Side–Angle Case Two triangles are congruent if two of the angles are equal and the lengths of the corresponding sides between the two angles are equal. For example, in Figure 21(a), the two triangles are congruent because two angles and the included side are equal. 2. Side–Side–Side Case Two triangles are congruent if the lengths of the corresponding sides of the triangles are equal. For example, in Figure 21(b), the two triangles are congruent because the three corresponding sides are all equal. 3. Side–Angle–Side Case Two triangles are congruent if the lengths of two corresponding sides are equal and the angles between the two sides are the same. For example, in Figure 21(c), the two triangles are congruent because two sides and the included angle are equal. 15 20 80 80 10 15 20 40 (a) 40 8 40 7 10 40 8 8 7 8 (b) (c) Figure 21 We contrast congruent triangles with similar triangles. DEFINITION In Words Two triangles are similar if they have the same shape, but (possibly) different sizes. Two triangles are similar if the corresponding angles are equal and the lengths of the corresponding sides are proportional. For example, the triangles in Figure 22 (on the next page) are similar because the corresponding angles are equal. In addition, the lengths of the corresponding sides are proportional because each side in the triangle on the right is twice as long as each corresponding side in the triangle on the left. That is, the ratio of the corresponding f d e sides is a constant: = = = 2. a c b 34 CHAPTER R Review 80 d 5 2a a 80 30 e 5 2b b 30 70 70 f 5 2c c Figure 22 Similar triangles It is not necessary to verify that all three angles are equal and all three sides are proportional to determine whether two triangles are congruent. Determining Similar Triangles 1. Angle–Angle Case Two triangles are similar if two of the corresponding angles are equal. For example, in Figure 23(a), the two triangles are similar because two angles are equal. 2. Side–Side–Side Case Two triangles are similar if the lengths of all three sides of each triangle are proportional. For example, in Figure 23(b), the two triangles are similar because 10 5 6 1 = = = 30 15 18 3 3. Side–Angle–Side Case Two triangles are similar if two corresponding sides are proportional and the angles between the two sides are equal. For example, in Figure 23(c), the two triangles are similar because 4 12 2 = = and the angles between the sides are equal. 6 18 3 15 80 80 30 5 10 18 120 120 6 35 35 18 12 6 4 (a) (b) (c) Figure 23 EX AMPLE 5 Using Similar Triangles Given that the triangles in Figure 24 are similar, find the missing length x and the angles A, B, and C. 60 6 3 30 x B C 5 90 Figure 24 A SECTION R.3 Geometry Essentials Solution 35 Because the triangles are similar, corresponding angles are equal. So A = 90°, B = 60°, and C = 30°. Also, the corresponding sides are proportional. That is, 3 6 = . We solve this equation for x. x 5 3 6 = x 5 3 6 5x # = 5x # Multiply both sides by 5x. x 5 Simplify. 3x = 30 Divide both sides by 3. x = 10 r The missing length is 10 units. Now Work PROBLEM 43 Proof of the Pythagorean Theorem Begin with a square, each side of length a + b. In this square, form four right triangles, each having legs equal in length to a and b. See Figure 25. All these triangles are congruent (two sides and their included angle are equal). As a result, the hypotenuse of each is the same, say c, and the pink shading in Figure 25 indicates a square with an area equal to c 2. b Area 1 = –2 ab a 1 Area = –2 ab a c c b Area = c 2 c b c 1 Area = –2 ab a 1 Area = –2 ab a b Figure 25 The area of the original square with sides a + b equals the sum of the areas of the 1 four triangles (each of area ab) plus the area of the square with side c. That is, 2 1 1 1 1 1a + b2 2 = ab + ab + ab + ab + c 2 2 2 2 2 a2 + 2ab + b2 = 2ab + c 2 a 2 + b2 = c 2 The proof is complete. x b a ■ Proof of the Converse of the Pythagorean Theorem Begin with two triangles: one a right triangle with legs a and b and the other a triangle with sides a, b, and c for which c 2 = a2 + b2. See Figure 26. By the Pythagorean Theorem, the length x of the third side of the first triangle is x 2 = a 2 + b2 (a) But c 2 = a2 + b2. Then, c b a 2 2 (b) c = a + b Figure 26 2 x2 = c 2 x = c The two triangles have the same sides and are therefore congruent. This means corresponding angles are equal, so the angle opposite side c of the second triangle equals 90°. The proof is complete. ■ 36 CHAPTER R Review R.3 Assess Your Understanding Concepts and Vocabulary 10. True or False The triangles shown are congruent. 1. A(n) triangle is one that contains an angle of 90 degrees. The longest side is called the . 10 2. For a triangle with base b and altitude h, a formula for the area A is 30 . 30 3. The formula for the circumference C of a circle of radius r is . 29 29 10 if corresponding angles are equal 4. Two triangles are and the lengths of the corresponding sides are proportional. 11. True or False The triangles shown are similar. 5. Which of the following is not a case for determining congruent triangles? (a) Angle–Side–Angle (b) Side–Angle–Side (c) Angle–Angle–Angle (d) Side-Side-Side 25 25 6. Choose the formula for the volume of a sphere of radius r. 4 4 (a) pr 2 (b) pr 3 (c) 4pr 3 (d) 4pr 2 3 3 100 7. True or False In a right triangle, the square of the length of the longest side equals the sum of the squares of the lengths of the other two sides. 100 12. True or False The triangles shown are similar. 8. True or False The triangle with sides of lengths 6, 8, and 10 is a right triangle. 9. True or False The surface area of a sphere of radius r is 4 2 pr . 3 4 3 120⬚ 120⬚ 2 3 Skill Building In Problems 13–18, the lengths of the legs of a right triangle are given. Find the hypotenuse. 13. a = 5, b = 12 14. a = 6, b = 8 15. a = 10, b = 24 16. a = 4, b = 3 17. a = 7, b = 24 18. a = 14, b = 48 In Problems 19–26, the lengths of the sides of a triangle are given. Determine which are right triangles. For those that are, identify the hypotenuse. 19. 3, 4, 5 20. 6, 8, 10 21. 4, 5, 6 22. 2, 2, 3 23. 7, 24, 25 24. 10, 24, 26 25. 6, 4, 3 26. 5, 4, 7 27. Find the area A of a rectangle with length 4 inches and width 2 inches. 28. Find the area A of a rectangle with length 9 centimeters and width 4 centimeters. 29. Find the area A of a triangle with height 4 inches and base 2 inches. 30. Find the area A of a triangle with height 9 centimeters and base 4 centimeters. 31. Find the area A and circumference C of a circle of radius 5 meters. 32. Find the area A and circumference C of a circle of radius 2 feet. 33. Find the volume V and surface area S of a closed rectangular box with length 8 feet, width 4 feet, and height 7 feet. 34. Find the volume V and surface area S of a closed rectangular box with length 9 inches, width 4 inches, and height 8 inches. 35. Find the volume V and surface area S of a sphere of radius 4 centimeters. 36. Find the volume V and surface area S of a sphere of radius 3 feet. 37. Find the volume V and surface area S of a closed right circular cylinder with radius 9 inches and height 8 inches. 38. Find the volume V and surface area S of a closed right circular cylinder with radius 8 inches and height 9 inches. SECTION R.3 Geometry Essentials 37 In Problems 39–42, find the area of the shaded region. 39. 40. 2 41. 2 42. 2 2 2 2 2 2 In Problems 43–46, the triangles in each pair are similar. Find the missing length x and the missing angles A, B, and C. 43. 60 44. 2 30 16 90 45. 75 4 10 125 50 20 95 12 30 46. 60 25 50 45 75 A A x B x A 6 5 30 A B 8 8 B C x B C x C C Applications and Extensions 47. How many feet has a wheel with a diameter of 16 inches traveled after four revolutions? 48. How many revolutions will a circular disk with a diameter of 4 feet have completed after it has rolled 20 feet? 51. Architecture A Norman window consists of a rectangle surmounted by a semicircle. Find the area of the Norman window shown in the illustration. How much wood frame is needed to enclose the window? 49. In the figure shown, ABCD is a square, with each side of length 6 feet. The width of the border (shaded portion) between the outer square EFGH and ABCD is 2 feet. Find the area of the border. E F A 6' B 6 ft 2 ft D 4' C H G 50. Refer to the figure. Square ABCD has an area of 100 square feet; square BEFG has an area of 16 square feet. What is the area of the triangle CGF ? A B D E F G C 52. Construction A circular swimming pool that is 20 feet in diameter is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck? 3' 20' 38 CHAPTER R Review 53. How Tall Is the Great Pyramid? The ancient Greek philosopher Thales of Miletus is reported on one occasion to have visited Egypt and calculated the height of the Great Pyramid of Cheops by means of shadow reckoning. Thales knew that each side of the base of the pyramid was 252 paces and that his own height was 2 paces. He measured the length of the pyramid’s shadow to be 114 paces and determined the length of his shadow to be 3 paces. See the illustration. Using similar triangles, determine the height of the Great Pyramid in terms of the number of paces. 54. The Bermuda Triangle Karen is doing research on the Bermuda Triangle which she defines roughly by Hamilton, Bermuda; San Juan, Puerto Rico; and Fort Lauderdale, Florida. On her atlas Karen measures the straight-line distances from Hamilton to Fort Lauderdale, Fort Lauderdale to San Juan, and San Juan to Hamilton to be approximately 57 millimeters (mm), 58 mm, and 53.5 mm respectively. If the actual distance from Fort Lauderdale to San Juan is 1046 miles, approximate the actual distances from San Juan to Hamilton and from Hamilton to Fort Lauderdale. Source: Diggins, Julie E, String Straightedge and Shadow: The Story of Geometry, 2003, Whole Spirit Press, http://wholespiritpress.com. In Problems 55–57, use the facts that the radius of Earth is 3960 miles and 1 mile = 5280 feet. 55. How Far Can You See? The conning tower of the U.S.S. Silversides, a World War II submarine now permanently stationed in Muskegon, Michigan, is approximately 20 feet above sea level. How far can you see from the conning tower? 56. How Far Can You See? A person who is 6 feet tall is standing on the beach in Fort Lauderdale, Florida, and looks out onto the Atlantic Ocean. Suddenly, a ship appears on the horizon. How far is the ship from shore? How far can a person see from the bridge, which is 150 feet above sea level? 58. Suppose that m and n are positive integers with m 7 n. If a = m2 - n2, b = 2mn, and c = m2 + n2, show that a, b, and c are the lengths of the sides of a right triangle. (This formula can be used to find the sides of a right triangle that are integers, such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triples.) 57. How Far Can You See? The deck of a destroyer is 100 feet above sea level. How far can a person see from the deck? Explaining Concepts: Discussion and Writing 59. You have 1000 feet of flexible pool siding and intend to construct a swimming pool. Experiment with rectangular-shaped pools with perimeters of 1000 feet. How do their areas vary? What is the shape of the rectangle with the largest area? Now compute the area enclosed by a circular pool with a perimeter (circumference) of 1000 feet. What would be your choice of shape for the pool? If rectangular, what is your preference for dimensions? Justify your choice. If your only consideration is to have a pool that encloses the most area, what shape should you use? 60. The Gibb’s Hill Lighthouse, Southampton, Bermuda, in operation since 1846, stands 117 feet high on a hill 245 feet high, so its beam of light is 362 feet above sea level. A brochure states that the light itself can be seen on the horizon about 26 miles distant. Verify the accuracy of this information. The brochure further states that ships 40 miles away can see the light and that planes flying at 10,000 feet can see it 120 miles away. Verify the accuracy of these statements. What assumption did the brochure make about the height of the ship? 120 miles 40 miles SECTION R.4 Polynomials 39 R.4 Polynomials OBJECTIVES 1 2 3 4 5 6 7 Recognize Monomials (p. 39) Recognize Polynomials (p. 40) Add and Subtract Polynomials (p. 41) Multiply Polynomials (p. 42) Know Formulas for Special Products (p. 43) Divide Polynomials Using Long Division (p. 44) Work with Polynomials in Two Variables (p. 47) We have described algebra as a generalization of arithmetic in which letters are used to represent real numbers. From now on, we shall use the letters at the end of the alphabet, such as x, y, and z, to represent variables and use the letters at the beginning of the alphabet, such as a, b, and c, to represent constants. In the expressions 3x + 5 and ax + b, it is understood that x is a variable and that a and b are constants, even though the constants a and b are unspecified. As you will find out, the context usually makes the intended meaning clear. 1 Recognize Monomials DEFINITION NOTE The nonnegative integers are the integers 0, 1, 2, 3,…. ■ A monomial in one variable is the product of a constant and a variable raised to a nonnegative integer power. A monomial is of the form axk where a is a constant, x is a variable, and k Ú 0 is an integer. The constant a is called the coefficient of the monomial. If a ≠ 0, then k is called the degree of the monomial. EXAMPL E 1 Examples of Monomials Monomial (a) (b) (c) (d) (e) EXAMPL E 2 2 6x - 22x3 3 - 5x x4 Coefficient Degree 6 - 22 3 -5 1 2 3 0 1 4 Since 3 = 3 # 1 = 3x 0, x ≠ 0 Since - 5x = - 5x 1 Since x 4 = 1 # x 4 r Examples of Nonmonomial Expressions 1 1 (a) 3x1>2 is not a monomial, since the exponent of the variable x is , and is not a 2 2 nonnegative integer. (b) 4x -3 is not a monomial, since the exponent of the variable x is - 3, and - 3 is not a nonnegative integer. r Now Work PROBLEM 9 40 CHAPTER R Review 2 Recognize Polynomials Two monomials with the same variable raised to the same power are called like terms. For example, 2x4 and - 5x4 are like terms. In contrast, the monomials 2x3 and 2x5 are not like terms. We can add or subtract like terms using the Distributive Property. For example, 2x2 + 5x2 = 12 + 52x2 = 7x2 and 8x3 - 5x3 = 18 - 52x3 = 3x3 The sum or difference of two monomials having different degrees is called a binomial. The sum or difference of three monomials with three different degrees is called a trinomial. For example, x2 - 2 is a binomial. x3 - 3x + 5 is a trinomial. 2x2 + 5x2 + 2 = 7x2 + 2 is a binomial. DEFINITION A polynomial in one variable is an algebraic expression of the form an xn + an - 1 xn - 1 + g + a1 x + a0 (1) where an , an - 1 ,c, a1 , a0 are constants,* called the coefficients of the polynomial, n Ú 0 is an integer, and x is a variable. If an ≠ 0, it is called the leading coefficient, anxn is called the leading term, and n is the degree of the polynomial. In Words A polynomial is a sum of monomials. The monomials that make up a polynomial are called its terms. If all of the coefficients are 0, the polynomial is called the zero polynomial, which has no degree. Polynomials are usually written in standard form, beginning with the nonzero term of highest degree and continuing with terms in descending order according to degree. If a power of x is missing, it is because its coefficient is zero. EX AMPLE 3 Examples of Polynomials Polynomial 3 2 - 8x + 4x - 6x + 2 Coefficients Degree 3x - 5 = 3x + 0 # x + 1 - 52 8 - 2x + x2 = 1 # x2 + 1 - 22x + 8 - 8, 4, - 6, 2 3 3, 0, - 5 2 1, - 2, 8 2 5x + 22 = 5x + 22 5, 22 1 3 = 3 # 1 = 3 # x0 3 0 0 0 No degree 2 2 1 r Although we have been using x to represent the variable, letters such as y and z are also commonly used. 3x4 - x2 + 2 is a polynomial (in x) of degree 4. 9y3 - 2y2 + y - 3 is a polynomial (in y) of degree 3. z5 + p is a polynomial (in z) of degree 5. Algebraic expressions such as 1 x and x2 + 1 x + 5 *The notation an is read as “a sub n.” The number n is called a subscript and should not be confused with an exponent. We use subscripts to distinguish one constant from another when a large or undetermined number of constants are required. SECTION R.4 Polynomials 41 1 = x -1 has an exponent x that is not a nonnegative integer. Although the second expression is the quotient of two polynomials, the polynomial in the denominator has degree greater than 0, so the expression cannot be a polynomial. are not polynomials. The first is not a polynomial because Now Work PROBLEM 19 3 Add and Subtract Polynomials Polynomials are added and subtracted by combining like terms. EXAMPL E 4 Adding Polynomials Find the sum of the polynomials: 8x3 - 2x2 + 6x - 2 and 3x4 - 2x3 + x2 + x Solution We shall find the sum in two ways. Horizontal Addition: The idea here is to group the like terms and then combine them. 18x3 - 2x2 + 6x - 22 + 13x4 - 2x3 + x2 + x2 = 3x4 + 18x3 - 2x3 2 + 1 - 2x2 + x2 2 + 16x + x2 - 2 = 3x4 + 6x3 - x2 + 7x - 2 Vertical Addition: The idea here is to vertically line up the like terms in each polynomial and then add the coefficients. x4 x3 x2 3 x1 x0 2 8x - 2x + 6x - 2 + 3x4 - 2x3 + x2 + x 3x4 + 6x3 - x2 + 7x - 2 We can subtract two polynomials horizontally or vertically as well. EXAMPL E 5 Solution Subtracting Polynomials Find the difference: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52 Horizontal Subtraction: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52 = 3x4 - 4x3 + 6x2 - 1 + ( - 2x4 + 8x2 + 6x - 5) y Be sure to change the sign of each term in the second polynomial. = 13x4 - 2x4 2 + 1 - 4x3 2 + 16x2 + 8x2 2 + 6x + 1 - 1 - 52 c Group like terms. = x4 - 4x3 + 14x2 + 6x - 6 r 42 CHAPTER R Review COMMENT Vertical subtraction will be used when we divide polynomials. ■ Vertical Subtraction: We line up like terms, change the sign of each coefficient of the second polynomial, and add. x4 x3 x1 x2 x0 x4 3x4 - 4x3 + 6x2 - 1 = 4 2 - 3 2x - 8x - 6x + 54 = + x3 x2 x1 x0 3x4 - 4x3 + 6x2 - 1 4 2 - 2x + 8x + 6x - 5 x4 - 4x3 + 14x2 + 6x - 6 r Which of these methods to use for adding and subtracting polynomials is up to you. To save space, we shall most often use the horizontal format. Now Work PROBLEM 31 4 Multiply Polynomials Two monomials may be multiplied using the Laws of Exponents and the Commutative and Associative Properties. For example, 12x3 2 # 15x4 2 = 12 # 52 # 1x3 # x4 2 = 10x3 + 4 = 10x7 Products of polynomials are found by repeated use of the Distributive Property and the Laws of Exponents. Again, you have a choice of horizontal or vertical format. EX AMPLE 6 Solution Multiplying Polynomials Find the product: 12x + 52 1x2 - x + 22 Horizontal Multiplication: 12x + 52 1x2 - x + 22 = 2x1x2 - x + 22 + 51x2 - x + 22 c Distributive Property = 12x # x2 - 2x # x + 2x # 22 + 15 # x2 - 5 # x + 5 # 22 c Distributive Property = 12x3 - 2x2 + 4x2 + 15x2 - 5x + 102 c Law of Exponents = 2x3 + 3x2 - x + 10 c Combine like terms. Vertical Multiplication: The idea here is very much like multiplying a two-digit number by a three-digit number. x2 - x 2x 2x3 - 2x2 + 4x 1+2 5x2 - 5x 3 2x + 3x2 - x Now Work PROBLEM + + 47 2 5 This line is 2x(x 2 - x + 2). + 10 + 10 This line is 5(x 2 - x + 2). Sum of the above two lines r SECTION R.4 Polynomials 43 5 Know Formulas for Special Products Certain products, which we call special products, occur frequently in algebra. We can calculate them easily using the FOIL (First, Outer, Inner, Last) method of multiplying two binomials. Outer First (ax + b)(cx + d ) = ax(cx + d ) + b(cx + d ) Inner First Last Outer 2 2 Inner 2 Last 2 = ax # cx + ax # d + b # cx + b # d = acx2 + adx + bcx + bd = acx2 + (ad + bc)x + bd EXAMPL E 7 Using FOIL (a) 1x - 32 1x + 32 = x2 + 3x - 3x - 9 = x2 - 9 F (b) (c) (d) (e) O I L 1x + 22 = 1x + 22 1x + 22 = x + 2x + 2x + 4 1x - 32 2 = 1x - 32 1x - 32 = x2 - 3x - 3x + 9 1x + 32 1x + 12 = x2 + x + 3x + 3 = x2 + 4x + 12x + 12 13x + 42 = 6x2 + 8x + 3x + 4 = 6x2 + 2 Now Work 2 PROBLEMS 49 AND = x2 + 4x + 4 = x2 - 6x + 9 3 11x + 4 r 57 Some products have been given special names because of their form. The following special products are based on Examples 7(a), (b), and (c). Difference of Two Squares 1x - a2 1x + a2 = x2 - a2 (2) Squares of Binomials, or Perfect Squares 1x + a2 2 = x2 + 2ax + a2 (3a) 1x - a2 2 = x2 - 2ax + a2 EXAMPL E 8 (3b) Using Special Product Formulas (a) 1x - 52 1x + 52 = x2 - 52 = x2 - 25 (b) 1x + 72 2 = x2 + 2 # 7 # x + 72 = x2 + 14x + 49 (c) 12x + 12 2 = 12x2 2 + 2 # 1 # 2x + 12 = 4x2 + 4x + 1 (d) 13x - 42 2 = 13x2 2 - 2 # 4 # 3x + 42 = 9x2 - 24x + 16 Now Work PROBLEMS 67, 69, AND Difference of two squares Square of a binomial Notice that we used 2x in place of x in formula (3a). Replace x by 3x in formula (3b). 71 Let’s look at some more examples that lead to general formulas. r 44 CHAPTER R Review EX AMPLE 9 Cubing a Binomial (a) 1x + 22 3 = 1x + 22 1x + 22 2 = 1x + 22 1x2 + 4x + 42 Formula (3a) = 1x3 + 4x2 + 4x2 + 12x2 + 8x + 82 = x3 + 6x2 + 12x + 8 (b) 1x - 12 3 = 1x - 12 1x - 12 2 = 1x - 12 1x2 - 2x + 12 Formula (3b) = 1x3 - 2x2 + x2 - 1x2 - 2x + 12 = x3 - 3x2 + 3x - 1 r Cubes of Binomials, or Perfect Cubes 1x + a2 3 = x3 + 3ax2 + 3a2 x + a3 1x - a2 3 = x3 - 3ax2 + 3a2 x - a3 Now Work EX A MPL E 1 0 PROBLEM (4a) (4b) 87 Forming the Difference of Two Cubes 1x - 12 1x2 + x + 12 = x1x2 + x + 12 - 11x2 + x + 12 = x3 + x2 + x - x2 - x - 1 r = x3 - 1 EX A MPL E 11 Forming the Sum of Two Cubes 1x + 22 1x2 - 2x + 42 = x1x2 - 2x + 42 + 21x2 - 2x + 42 = x3 - 2x2 + 4x + 2x2 - 4x + 8 r = x3 + 8 Examples 10 and 11 lead to two more special products. Difference of Two Cubes 1x - a2 1x2 + ax + a2 2 = x3 - a3 (5) Sum of Two Cubes 1x + a2 1x2 - ax + a2 2 = x3 + a3 (6) 6 Divide Polynomials Using Long Division The procedure for dividing two polynomials is similar to the procedure for dividing two integers. SECTION R.4 Polynomials EX AM PL E 12 45 Dividing Two Integers Divide 842 by 15. Solution 56 Divisor S 15 ) 842 75 92 90 2 So, d Quotient d Dividend d 5 # 15 (subtract) d 6 # 15 (subtract) d Remainder 842 2 = 56 + . 15 15 r In the long-division process detailed in Example 12, the number 15 is called the divisor, the number 842 is called the dividend, the number 56 is called the quotient, and the number 2 is called the remainder. To check the answer obtained in a division problem, multiply the quotient by the divisor and add the remainder. The answer should be the dividend. 1Quotient2 1Divisor2 + Remainder = Dividend For example, we can check the results obtained in Example 12 as follows: 1562 1152 + 2 = 840 + 2 = 842 To divide two polynomials, we first must write each polynomial in standard form. The process then follows a pattern similar to that of Example 12. The next example illustrates the procedure. EX AM PL E 1 3 Dividing Two Polynomials Find the quotient and the remainder when 3x3 + 4x2 + x + 7 is divided by x2 + 1 Solution Each polynomial is in standard form. The dividend is 3x3 + 4x2 + x + 7, and the divisor is x2 + 1. NOTE Remember, a polynomial is in standard form when its terms are written in descending powers of x. ■ STEP 1: Divide the leading term of the dividend, 3x3, by the leading term of the divisor, x2. Enter the result, 3x, over the term 3x3, as follows: 3x x + 1 ) 3x3 + 4x2 + x + 7 2 STEP 2: Multiply 3x by x2 + 1, and enter the result below the dividend. 3x x2 + 1 ) 3x3 + 4x2 + x + 7 d 3x # (x 2 + 1) = 3x 3 + 3x 3x3 + 3x c Align the 3x term under the x to make the next step easier. STEP 3: Subtract and bring down the remaining terms. 3x x + 1 ) 3x3 + 4x2 + x + 7 d Subtract (change the signs and add). 3x3 + 3x 2 4x - 2x + 7 d Bring down the 4x 2 and the 7. 2 46 CHAPTER R Review STEP 4: Repeat Steps 1–3 using 4x2 - 2x + 7 as the dividend. 3x + 4 x + 1 ) 3x3 + 4x2 + x 3x3 + 3x 4x2 - 2x 4x2 - 2x 2 COMMENT If the degree of the divisor is greater than the degree of the dividend, then the process ends. ■ d + 7 + 7 + 4 + 3 d Divide 4x 2 by x 2 to get 4. d Multiply x 2 + 1 by 4; subtract. Since x2 does not divide - 2x evenly (that is, the result is not a monomial), the process ends. The quotient is 3x + 4, and the remainder is - 2x + 3. Check: 1Quotient2 1Divisor2 + Remainder = 13x + 42 1x2 + 12 + 1 - 2x + 32 = 3x3 + 3x + 4x2 + 4 + 1 - 2x + 32 = 3x3 + 4x2 + x + 7 = Dividend Then 3x3 + 4x2 + x + 7 - 2x + 3 = 3x + 4 + 2 2 x + 1 x + 1 r The next example combines the steps involved in long division. EX A MPL E 1 4 Dividing Two Polynomials Find the quotient and the remainder when x4 - 3x3 + 2x - 5 is divided by x2 - x + 1 Solution In setting up this division problem, it is necessary to leave a space for the missing x2 term in the dividend. Divisor S Subtract S Subtract S Subtract S x2 - 2x - 3 x2 - x + 1 ) x4 - 3x3 x4 - x3 + x2 - 2x3 - x2 - 2x3 + 2x2 - 3x2 - 3x2 + 2x - 5 + + + 2x 2x 4x 3x x d Quotient d Dividend - 5 - 5 - 3 - 2 d Remainder Check: 1Quotient2 1Divisor2 + Remainder = 1x2 - 2x - 32 1x2 - x + 12 + x - 2 = x4 - x3 + x2 - 2x3 + 2x2 - 2x - 3x2 + 3x - 3 + x - 2 = x4 - 3x3 + 2x - 5 = Dividend As a result, x - 2 x4 - 3x3 + 2x - 5 = x2 - 2x - 3 + 2 x2 - x + 1 x - x + 1 r SECTION R.4 Polynomials 47 The process of dividing two polynomials leads to the following result: THEOREM Let Q be a polynomial of positive degree, and let P be a polynomial whose degree is greater than or equal to the degree of Q. The remainder after dividing P by Q is either the zero polynomial or a polynomial whose degree is less than the degree of the divisor Q. Now Work PROBLEM 95 7 Work with Polynomials in Two Variables A monomial in two variables x and y has the form axn ym, where a is a constant, x and y are variables, and n and m are nonnegative integers. The degree of a monomial is the sum of the powers of the variables. For example, 2xy3, x2 y2, and x3 y are all monomials that have degree 4. A polynomial in two variables x and y is the sum of one or more monomials in two variables. The degree of a polynomial in two variables is the highest degree of all the monomials with nonzero coefficients. Examples of Polynomials in Two Variables EX AM PL E 1 5 px3 - y2 3x2 + 2x3 y + 5 Two variables, degree is 4. Two variables, degree is 3. x4 + 4x3 y - xy3 + y4 Two variables, degree is 4. r Multiplying polynomials in two variables is handled in the same way as multiplying polynomials in one variable. Using a Special Product Formula EX AM PL E 1 6 To multiply 12x - y2 2, use the Squares of Binomials formula (3b) with 2x instead of x and with y instead of a. 12x - y2 2 = 12x2 2 - 2 # y # 2x + y2 = 4x2 - 4xy + y2 Now Work PROBLEM r 81 R.4 Assess Your Understanding Concepts and Vocabulary 1. The polynomial 3x4 - 2x3 + 13x2 - 5 is of degree leading coefficient is . 2. 1x2 - 42 1x2 + 42 = 3. 1x - 22 1x2 + 2x + 42 = . The 5. Choose the degree of the monomial 3x4 y2. (a) 3 (b) 8 (c) 6 (d) 2 6. True or False 4x -2 is a monomial of degree - 2. . . 4. The monomials that make up a polynomial are called which of the following? (a) terms (b) variables (c) factors (d) coefficients 7. True or False The degree of the product of two nonzero polynomials equals the sum of their degrees. 8. True or False 1x + a2 1x2 + ax + a2 = x3 + a3. 48 CHAPTER R Review Skill Building In Problems 9–18, tell whether the expression is a monomial. If it is, name the variable(s) and the coefficient, and give the degree of the monomial. If it is not a monomial, state why not. 9. 2x3 14. 5x2 y3 10. - 4x2 15. 11. 8x y 8 x 16. - 2x2 y3 12. - 2x -3 13. - 2xy2 17. x2 + y2 18. 3x2 + 4 In Problems 19–28, tell whether the expression is a polynomial. If it is, give its degree. If it is not, state why not. 19. 3x2 - 5 23. 3x2 - 22. - p 21. 5 20. 1 - 4x 5 x x2 + 5 3x3 + 2x - 1 28. x3 - 1 x2 + x + 1 In Problems 29–48, add, subtract, or multiply, as indicated. Express your answer as a single polynomial in standard form. 24. 3 + 2 x 25. 2y3 - 22 26. 10z2 + z 27. 29. 1x2 + 4x + 52 + 13x - 32 30. 1x3 + 3x2 + 22 + 1x2 - 4x + 42 31. 1x3 - 2x2 + 5x + 102 - 12x2 - 4x + 32 32. 1x2 - 3x - 42 - 1x3 - 3x2 + x + 52 33. 16x5 + x3 + x2 + 15x4 - x3 + 3x2 2 34. 110x5 - 8x2 2 + 13x3 - 2x2 + 62 35. 1x2 - 3x + 12 + 213x2 + x - 42 36. - 21x2 + x + 12 + 1 - 5x2 - x + 22 37. 61x3 + x2 - 32 - 412x3 - 3x2 2 38. 814x3 - 3x2 - 12 - 614x3 + 8x - 22 39. 1x2 - x + 22 + 12x2 - 3x + 52 - 1x2 + 12 40. 1x2 + 12 - 14x2 + 52 + 1x2 + x - 22 41. 91y2 - 3y + 42 - 611 - y2 2 42. 811 - y3 2 + 411 + y + y2 + y3 2 43. x1x2 + x - 42 44. 4x2 1x3 - x + 22 45. - 2x2 14x3 + 52 46. 5x3 13x - 42 47. 1x + 12 1x2 + 2x - 42 48. 12x - 32 1x2 + x + 12 In Problems 49–66, multiply the polynomials using the FOIL method. Express your answer as a single polynomial in standard form. 49. 1x + 22 1x + 42 50. 1x + 32 1x + 52 51. 12x + 52 1x + 22 52. 13x + 12 12x + 12 53. 1x - 42 1x + 22 54. 1x + 42 1x - 22 55. 1x - 32 1x - 22 56. 1x - 52 1x - 12 57. 12x + 32 1x - 22 58. 12x - 42 13x + 12 59. 1 - 2x + 32 1x - 42 60. 1 - 3x - 12 1x + 12 61. 1 - x - 22 1 - 2x - 42 62. 1 - 2x - 32 13 - x2 63. 1x - 2y2 1x + y2 64. 12x + 3y2 1x - y2 65. 1 - 2x - 3y2 13x + 2y2 66. 1x - 3y2 1 - 2x + y2 In Problems 67–90, multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form. 67. 1x - 72 1x + 72 68. 1x - 12 1x + 12 69. 12x + 32 12x - 32 70. 13x + 22 13x - 22 71. 1x + 42 2 72. 1x + 52 2 73. 1x - 42 2 74. 1x - 52 2 75. 13x + 42 13x - 42 76. 15x - 32 15x + 32 77. 12x - 32 2 78. 13x - 42 2 SECTION R.5 Factoring Polynomials 79. 1x + y2 1x - y2 80. 1x + 3y2 1x - 3y2 81. 13x + y2 13x - y2 82. 13x + 4y2 13x - 4y2 83. 1x + y2 2 84. 1x - y2 2 85. 1x - 2y2 2 86. 12x + 3y2 2 87. 1x - 22 3 88. 1x + 12 3 89. 12x + 12 3 90. 13x - 22 3 49 In Problems 91–106, find the quotient and the remainder. Check your work by verifying that 1Quotient2 1Divisor2 + Remainder = Dividend 91. 4x3 - 3x2 + x + 1 divided by x + 2 92. 3x3 - x2 + x - 2 divided by x + 2 93. 4x3 - 3x2 + x + 1 divided by x2 94. 3x3 - x2 + x - 2 divided by x2 95. 5x4 - 3x2 + x + 1 divided by x2 + 2 96. 5x4 - x2 + x - 2 divided by x2 + 2 97. 4x5 - 3x2 + x + 1 divided by 2x3 - 1 98. 3x5 - x2 + x - 2 divided by 3x3 - 1 99. 2x4 - 3x3 + x + 1 divided by 2x2 + x + 1 100. 3x4 - x3 + x - 2 divided by 3x2 + x + 1 101. - 4x3 + x2 - 4 divided by x - 1 102. - 3x4 - 2x - 1 divided by x - 1 103. 1 - x2 + x4 divided by x2 + x + 1 104. 1 - x2 + x4 divided by x2 - x + 1 105. x3 - a3 divided by x - a 106. x5 - a5 divided by x - a Explaining Concepts: Discussion and Writing 107. Explain why the degree of the product of two nonzero polynomials equals the sum of their degrees. 108. Explain why the degree of the sum of two polynomials of different degrees equals the larger of their degrees. 109. Give a careful statement about the degree of the sum of two polynomials of the same degree. 110. Do you prefer adding two polynomials using the horizontal method or the vertical method? Write a brief position paper defending your choice. 111. Do you prefer to memorize the rule for the square of a binomial 1x + a2 2 or to use FOIL to obtain the product? Write a brief position paper defending your choice. R.5 Factoring Polynomials OBJECTIVES 1 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes (p. 50) 2 Factor Perfect Squares (p. 51) 3 Factor a Second-Degree Polynomial: x2 + Bx + C (p. 52) 4 Factor by Grouping (p. 53) 5 Factor a Second-Degree Polynomial: Ax2 + Bx + C, A ≠ 1 (p. 54) 6 Complete the Square (p. 56) Consider the following product: 12x + 32 1x - 42 = 2x2 - 5x - 12 The two polynomials on the left side are called factors of the polynomial on the right side. Expressing a given polynomial as a product of other polynomials—that is, finding the factors of a polynomial—is called factoring. 50 CHAPTER R Review We shall restrict our discussion here to factoring polynomials in one variable into products of polynomials in one variable, where all coefficients are integers. We call this factoring over the integers. Any polynomial can be written as the product of 1 times itself or as - 1 times its additive inverse. If a polynomial cannot be written as the product of two other polynomials (excluding 1 and - 1), then the polynomial is prime. When a polynomial has been written as a product consisting only of prime factors, it is factored completely. Examples of prime polynomials (over the integers) are COMMENT Over the real numbers, 3x + 4 factors into 3 1x + 43 2 . It is the noninteger 43 that causes 3x + 4 to be prime over the integers. In most instances, we will be factoring over the integers. ■ EX AMPLE 1 2, 3, 5, x, x + 1, x - 1, 3x + 4, x2 + 4 The first factor to look for in a factoring problem is a common monomial factor present in each term of the polynomial. If one is present, use the Distributive Property to factor it out. Continue factoring out monomial factors until none are left. Identifying Common Monomial Factors Polynomial Common Monomial Factor Remaining Factor Factored Form 2x + 4 2 x + 2 2x + 4 = 21x + 22 3x - 6 3 x - 2 3x - 6 = 31x - 22 2x - 4x + 8 2 x - 2x + 4 2x2 - 4x + 8 = 21x2 - 2x + 42 8x - 12 4 2x - 3 8x - 12 = 412x - 32 x + x x x + 1 x2 + x = x1x + 12 x3 - 3x2 x2 x - 3 x3 - 3x2 = x2 1x - 32 6x2 + 9x 3x 2x + 3 6x2 + 9x = 3x12x + 32 2 2 2 r Notice that once all common monomial factors have been removed from a polynomial, the remaining factor is either a prime polynomial of degree 1 or a polynomial of degree 2 or higher. (Do you see why?) Now Work PROBLEM 9 1 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes When you factor a polynomial, first check for common monomial factors. Then see whether you can use one of the special formulas discussed in the previous section. Difference of Two Squares Perfect Squares Sum of Two Cubes Difference of Two Cubes EX AMPLE 2 x2 x2 x2 x3 x3 + + - a2 = 1x - a2 1x + a2 2ax + a2 = 1x + a2 2 2ax + a2 = 1x - a2 2 a3 = 1x + a2 1x2 - ax + a2 2 a3 = 1x - a2 1x2 + ax + a2 2 Factoring the Difference of Two Squares Factor completely: x2 - 4 Solution Note that x2 - 4 is the difference of two squares, x2 and 22. x2 - 4 = 1x - 22 1x + 22 r SECTION R.5 Factoring Polynomials EXAMPL E 3 51 Factoring the Difference of Two Cubes Factor completely: x3 - 1 Solution EXAMPL E 4 Because x3 - 1 is the difference of two cubes, x3 and 13, x3 - 1 = 1x - 12 1x2 + x + 12 r Factoring the Sum of Two Cubes Factor completely: x3 + 8 Solution EXAMPL E 5 Because x3 + 8 is the sum of two cubes, x3 and 23, x3 + 8 = 1x + 22 1x2 - 2x + 42 r Factoring the Difference of Two Squares Factor completely: x4 - 16 Solution Because x4 - 16 is the difference of two squares, x4 = 1x2 2 and 16 = 42, 2 x4 - 16 = 1x2 - 42 1x2 + 42 But x2 - 4 is also the difference of two squares. Then, x4 - 16 = 1x2 - 42 1x2 + 42 = 1x - 22 1x + 22 1x2 + 42 Now Work PROBLEMS 19 AND r 37 2 Factor Perfect Squares When the first term and third term of a trinomial are both positive and are perfect squares, such as x2, 9x2, 1, and 4, check to see whether the trinomial is a perfect square. EXAMPL E 6 Factoring a Perfect Square Factor completely: x2 + 6x + 9 Solution The first term, x2, and the third term, 9 = 32, are perfect squares. Because the middle term, 6x, is twice the product of x and 3, we have a perfect square. x2 + 6x + 9 = 1x + 32 2 EXAMPL E 7 r Factoring a Perfect Square Factor completely: 9x2 - 6x + 1 Solution The first term, 9x2 = 13x2 2, and the third term, 1 = 12, are perfect squares. Because the middle term, - 6x, is - 2 times the product of 3x and 1, we have a perfect square. 9x2 - 6x + 1 = 13x - 12 2 EXAMPL E 8 r Factoring a Perfect Square Factor completely: 25x2 + 30x + 9 Solution The first term, 25x2 = 15x2 2, and the third term, 9 = 32, are perfect squares. Because the middle term, 30x, is twice the product of 5x and 3, we have a perfect square. 25x2 + 30x + 9 = 15x + 32 2 Now Work PROBLEMS 29 AND 103 r 52 CHAPTER R Review If a trinomial is not a perfect square, it may be possible to factor it using the technique discussed next. 3 Factor a Second-Degree Polynomial: x2 + Bx + C The idea behind factoring a second-degree polynomial like x2 + Bx + C is to see whether it can be made equal to the product of two (possibly equal) first-degree polynomials. For example, consider 1x + 32 1x + 42 = x2 + 7x + 12 The factors of x2 + 7x + 12 are x + 3 and x + 4. Notice the following: x2 + 7x + 12 = (x + 3)(x + 4) 12 is the product of 3 and 4 7 is the sum of 3 and 4 In general, if x2 + Bx + C = 1x + a2 1x + b2 = x2 + (a + b)x + ab, then ab = C and a + b = B. To factor a second-degree polynomial x2 + Bx + C, find integers whose product is C and whose sum is B. That is, if there are numbers a, b, where ab = C and a + b = B, then x2 + Bx + C = 1x + a2 1x + b2 EX AMPLE 9 Factoring a Trinomial Factor completely: x2 + 7x + 10 Solution First determine all pairs of integers whose product is 10, and then compute their sums. Integers whose product is 10 1, 10 - 1, - 10 2, 5 - 2, - 5 Sum 11 - 11 7 -7 The integers 2 and 5 have a product of 10 and add up to 7, the coefficient of the middle term. As a result, x2 + 7x + 10 = 1x + 22 1x + 52 EX A MPL E 1 0 r Factoring a Trinomial Factor completely: x2 - 6x + 8 Solution First determine all pairs of integers whose product is 8, and then compute each sum. Integers whose product is 8 1, 8 - 1, - 8 2, 4 - 2, - 4 Sum 9 -9 6 -6 Since - 6 is the coefficient of the middle term, x2 - 6x + 8 = 1x - 22 1x - 42 r SECTION R.5 Factoring Polynomials EX AM PL E 11 53 Factoring a Trinomial Factor completely: x2 - x - 12 Solution First determine all pairs of integers whose product is - 12, and then compute each sum. Integers whose product is − 12 1, - 12 - 1, 12 2, - 6 - 2, 6 3, - 4 - 3, 4 Sum - 11 11 -4 4 -1 1 Since - 1 is the coefficient of the middle term, x2 - x - 12 = 1x + 32 1x - 42 EX AM PL E 1 2 r Factoring a Trinomial Factor completely: x2 + 4x - 12 Solution The integers - 2 and 6 have a product of - 12 and have the sum 4. So, x2 + 4x - 12 = 1x - 22 1x + 62 r To avoid errors in factoring, always check your answer by multiplying it out to see whether the result equals the original expression. When none of the possibilities works, the polynomial is prime. EX AM PL E 1 3 Identifying a Prime Polynomial Show that x2 + 9 is prime. Solution First list the pairs of integers whose product is 9, and then compute their sums. Integers whose product is 9 1, 9 - 1, - 9 3, 3 - 3, - 3 Sum 10 - 10 6 -6 Since the coefficient of the middle term in x2 + 9 = x2 + 0x + 9 is 0 and none of the sums equals 0, we conclude that x2 + 9 is prime. r Example 13 demonstrates a more general result: THEOREM Any polynomial of the form x2 + a2, a real, is prime. Now Work PROBLEMS 43 AND 87 4 Factor by Grouping Sometimes a common factor does not occur in every term of the polynomial but does occur in each of several groups of terms that together make up the polynomial. When this happens, the common factor can be factored out of each group by means of the Distributive Property. This technique is called factoring by grouping. EX AM PL E 1 4 Solution Factoring by Grouping Factor completely by grouping: 1x2 + 22x + 1x2 + 22 # 3 Notice the common factor x2 + 2. Applying the Distributive Property yields 1x2 + 22x + 1x2 + 22 # 3 = 1x2 + 22 1x + 32 Since x2 + 2 and x + 3 are prime, the factorization is complete. r 54 CHAPTER R Review The next example shows a factoring problem that occurs in calculus. EX A MPL E 1 5 Solution Factoring by Grouping Factor completely by grouping: 31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3 Here, 1x - 12 2 1x + 22 3 is a common factor of both 31x - 12 2 1x + 22 4 and 41x - 12 3 1x + 22 3. As a result, 31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3 = 1x - 12 2 1x + 22 3 3 31x + 22 + 41x - 12 4 = 1x - 12 2 1x + 22 3 3 3x + 6 + 4x - 44 = 1x - 12 2 1x + 22 3 17x + 22 EX A MPL E 1 6 r Factoring by Grouping Factor completely by grouping: x3 - 4x2 + 2x - 8 Solution To see whether factoring by grouping will work, group the first two terms and the last two terms. Then look for a common factor in each group. In this example, factor x2 from x3 - 4x2 and 2 from 2x - 8. The remaining factor in each case is the same, x - 4. This means that factoring by grouping will work, as follows: x3 - 4x2 + 2x - 8 = 1x3 - 4x2 2 + 12x - 82 = x2 1x - 42 + 21x - 42 = 1x - 42 1x2 + 22 Since x2 + 2 and x - 4 are prime, the factorization is complete. Now Work PROBLEMS 55 AND r 131 5 Factor a Second-Degree Polynomial: Ax2 + Bx + C, A 3 1 To factor a second-degree polynomial Ax2 + Bx + C, when A ≠ 1 and A, B, and C have no common factors, follow these steps: Steps for Factoring Ax2 + Bx + C, When A 3 1 and A, B, and C Have No Common Factors STEP 1: STEP 2: STEP 3: STEP 4: EX AMPLE 17 Find the value of AC. Find a pair of integers whose product is AC and that add up to B. That is, find a and b such that ab = AC and a + b = B. Write Ax2 + Bx + C = Ax2 + ax + bx + C. Factor this last expression by grouping. Factoring a Trinomial Factor completely: 2x2 + 5x + 3 Solution Comparing 2x2 + 5x + 3 to Ax2 + Bx + C, we find that A = 2, B = 5, and C = 3. STEP 1: The value of AC is 2 # 3 = 6. STEP 2: Determine the pairs of integers whose product is AC = 6 and compute their sums. Integers whose product is 6 1, 6 - 1, - 6 2, 3 - 2, - 3 Sum 7 -7 5 -5 SECTION R.5 Factoring Polynomials 55 STEP 3: The integers whose product is 6 that add up to B = 5 are 2 and 3. 2x2 + 5x + 3 = 2x2 + 2x + 3x + 3 STEP 4: Factor by grouping. 2x2 + 2x + 3x + 3 = 12x2 + 2x2 + 13x + 32 = 2x1x + 12 + 31x + 12 = 1x + 12 12x + 32 As a result, EX AMPL E 1 8 2x2 + 5x + 3 = 1x + 12 12x + 32 r Factoring a Trinomial Factor completely: 2x2 - x - 6 Solution Comparing 2x2 - x - 6 to Ax2 + Bx + C, we find that A = 2, B = - 1, and C = - 6. STEP 1: The value of AC is 2 # 1 - 62 = - 12. STEP 2: Determine the pairs of integers whose product is AC = - 12 and compute their sums. Integers whose product is − 12 1, - 12 - 1, 12 2, - 6 - 2, 6 3, - 4 - 3, 4 Sum - 11 11 -4 4 -1 1 STEP 3: The integers whose product is - 12 that add up to B = - 1 are - 4 and 3. 2x2 - x - 6 = 2x2 - 4x + 3x - 6 STEP 4: Factor by grouping. 2x2 - 4x + 3x - 6 = 12x2 - 4x2 + 13x - 62 = 2x1x - 22 + 31x - 22 = 1x - 22 12x + 32 As a result, Now Work 2x2 - x - 6 = 1x - 22 12x + 32 PROBLEM 61 SUMMARY Type of Polynomial Method Example Any polynomial Look for common monomial factors. (Always do this first!) 6x2 + 9x = 3x12x + 32 Binomials of degree 2 or higher Check for a special product: Difference of two squares, x2 - a2 Difference of two cubes, x3 - a3 Sum of two cubes, x3 + a3 x2 - 16 = 1x - 42 1x + 42 x3 - 64 = 1x - 42 1x2 + 4x + 162 x3 + 27 = 1x + 32 1x2 - 3x + 92 Trinomials of degree 2 Four or more terms Check for a perfect square, 1x { a2 2 Factoring x2 + Bx + C (p. 52) Factoring Ax2 + Bx + C (p. 54) x2 + 8x + 16 = 1x + 42 2 x2 - 10x + 25 = 1x - 52 2 x2 - x - 2 = 1x - 22 1x + 12 6x2 + x - 1 = 12x + 12 13x - 12 Grouping 2x3 - 3x2 + 4x - 6 = 12x - 32 1x2 + 22 r 56 CHAPTER R Review 6 Complete the Square The idea behind completing the square in one variable is to “adjust” an expression of the form x2 + bx to make it a perfect square. Perfect squares are trinomials of the form x2 + 2ax + a2 = (x + a)2 or x2 - 2ax + a2 = (x - a)2 For example, x2 + 6x + 9 is a perfect square because x2 + 6x + 9 = (x + 3)2. And p2 - 12p + 36 is a perfect square because p2 - 12p + 36 = (p - 6)2. So how do we “adjust” x2 + bx to make it a perfect square? We do it by adding a number. For example, to make x2 + 6x a perfect square, add 9. But how do we know to add 9? If we divide the coefficient of the first-degree term, 6, by 2, and then square the result, we obtain 9. This approach works in general. 1 2 WARNING To use a bb to complete 2 the square, the coefficient of the x 2 term must be 1. ■ EX A MPL E 1 9 Completing the Square of x 2 + bx 1 2 and then square the result. That is, determine the value of b in x2 + bx and 1 2 compute a b b . 2 Identify the coefficient of the first-degree term. Multiply this coefficient by Completing the Square Determine the number that must be added to each expression to complete the square. Then factor the expression. Start Add Result Factored Form y2 + 8y a 2 1# 8b = 16 2 y2 + 8y + 16 (y + 4)2 x2 + 12x a 2 1# 12b = 36 2 x2 + 12x + 36 (x + 6)2 a2 - 20a a 2 1# ( - 20) b = 100 2 a2 - 20a + 100 (a - 10)2 p2 - 5p a 2 1# 25 ( - 5) b = 2 4 p2 - 5p + 25 4 ap - 5 2 b 2 r Notice that the factored form of a perfect square is either y y 4 Area 5 y 2 Area 5 4y Figure 27 4 Area 5 4y b 2 b 2 b 2 b 2 x2 + bx + a b = ax + b or x2 - bx + a b = ax - b 2 2 2 2 Now Work PROBLEM 73 Are you wondering why we refer to making an expression a perfect square as “completing the square”? Look at the square in Figure 27. Its area is (y + 4)2. The yellow area is y2 and each orange area is 4y (for a total area of 8y). The sum of these areas is y2 + 8y. To complete the square, we need to add the area of the green region: 4 # 4 = 16. As a result, y2 + 8y + 16 = (y + 4)2. SECTION R.5 Factoring Polynomials 57 R.5 Assess Your Understanding Concepts and Vocabulary 1. If factored completely, 3x3 - 12x = 4. Choose the best description of x2 - 64. (a) Prime (b) Difference of two squares (c) Difference of two cubes (d) Perfect Square . 2. If a polynomial cannot be written as the product of two other polynomials (excluding 1 and - 1), then the polynomial is said to be . 5. Choose the complete factorization of 4x2 - 8x - 60. (a) 21x + 32 1x - 52 (b) 41x2 - 2x - 152 (c) 12x + 62 12x - 102 (d) 41x + 32 1x - 52 3. For x + Bx + C = 1x + a2 1x + b2 , which of the following must be true? (a) ab = B and a + b = C (b) a + b = C and a - b = B (c) ab = C and a + b = B (d) ab = B and a - b = C 2 6. To complete the square of x2 + bx, use which of the following? 1 2 1 (a) 12b2 2 (b) 2b2 (c) a bb (d) b2 2 2 7. True or False The polynomial x2 + 4 is prime. 8. True or False 3x3 - 2x2 - 6x + 4 = 13x - 22 1x2 + 22. Skill Building In Problems 9–18, factor each polynomial by removing the common monomial factor. 9. 3x + 6 11. ax2 + a 12. ax - a 10. 7x - 14 15. 2x2 - 2x 14. x3 - x2 + x 16. 3x2 - 3x 13. x3 + x2 + x 17. 3x2 y - 6xy2 + 12xy 18. 60x2 y - 48xy2 + 72x3 y In Problems 19–26, factor the difference of two squares. 19. x2 - 1 20. x2 - 4 21. 4x2 - 1 22. 9x2 - 1 23. x2 - 16 24. x2 - 25 25. 25x2 - 4 26. 36x2 - 9 In Problems 27–36, factor the perfect squares. 27. x2 + 2x + 1 28. x2 - 4x + 4 29. x2 + 4x + 4 30. x2 - 2x + 1 31. x2 - 10x + 25 32. x2 + 10x + 25 33. 4x2 + 4x + 1 34. 9x2 + 6x + 1 35. 16x2 + 8x + 1 36. 25x2 + 10x + 1 In Problems 37–42, factor the sum or difference of two cubes. 37. x3 - 27 38. x3 + 125 39. x3 + 27 40. 27 - 8x3 41. 8x3 + 27 42. 64 - 27x3 In Problems 43–54, factor each polynomial. 43. x2 + 5x + 6 44. x2 + 6x + 8 45. x2 + 7x + 6 46. x2 + 9x + 8 47. x2 + 7x + 10 48. x2 + 11x + 10 49. x2 - 10x + 16 50. x2 - 17x + 16 51. x2 - 7x - 8 52. x2 - 2x - 8 53. x2 + 7x - 8 54. x2 + 2x - 8 In Problems 55–60, factor by grouping. 55. 2x2 + 4x + 3x + 6 56. 3x2 - 3x + 2x - 2 57. 2x2 - 4x + x - 2 58. 3x2 + 6x - x - 2 59. 6x2 + 9x + 4x + 6 60. 9x2 - 6x + 3x - 2 In Problems 61–72, factor each polynomial. 61. 3x2 + 4x + 1 62. 2x2 + 3x + 1 63. 2z2 + 5z + 3 64. 6z2 + 5z + 1 65. 3x2 + 2x - 8 66. 3x2 + 10x + 8 67. 3x2 - 2x - 8 68. 3x2 - 10x + 8 69. 3x2 + 14x + 8 70. 3x2 - 14x + 8 71. 3x2 + 10x - 8 72. 3x2 - 10x - 8 In Problems 73–78, determine what number should be added to complete the square of each expression. Then factor each expression. 73. x2 + 10x 74. p2 + 14p 76. x2 - 4x 77. x2 - 1 x 2 75. y2 - 6y 78. x2 + 1 x 3 58 CHAPTER R Review Mixed Practice In Problems 79–126, factor each polynomial completely. If the polynomial cannot be factored, say it is prime. 79. x2 - 36 80. x2 - 9 81. 2 - 8x2 82. 3 - 27x2 83. x2 + 11x + 10 84. x2 + 5x + 4 85. x2 - 10x + 21 86. x2 - 6x + 8 87. 4x2 - 8x + 32 88. 3x2 - 12x + 15 89. x2 + 4x + 16 90. x2 + 12x + 36 91. 15 + 2x - x2 92. 14 + 6x - x2 93. 3x2 - 12x - 36 94. x3 + 8x2 - 20x 95. y4 + 11y3 + 30y2 96. 3y3 - 18y2 - 48y 97. 4x2 + 12x + 9 98. 9x2 - 12x + 4 99. 6x2 + 8x + 2 100. 8x2 + 6x - 2 101. x4 - 81 102. x4 - 1 103. x6 - 2x3 + 1 104. x6 + 2x3 + 1 105. x7 - x5 106. x8 - x5 107. 16x2 + 24x + 9 108. 9x2 - 24x + 16 109. 5 + 16x - 16x2 110. 5 + 11x - 16x2 111. 4y2 - 16y + 15 112. 9y2 + 9y - 4 113. 1 - 8x2 - 9x4 114. 4 - 14x2 - 8x4 115. x1x + 32 - 61x + 32 116. 513x - 72 + x13x - 72 117. 1x + 22 2 - 51x + 22 118. 1x - 12 2 - 21x - 12 119. 13x - 22 3 - 27 120. 15x + 12 3 - 1 121. 31x2 + 10x + 252 - 41x + 52 122. 71x2 - 6x + 92 + 51x - 32 123. x3 + 2x2 - x - 2 124. x3 - 3x2 - x + 3 125. x4 - x3 + x - 1 126. x4 + x3 + x + 1 Applications and Extensions In Problems 127–136, expressions that occur in calculus are given. Factor each expression completely. 127. 213x + 42 2 + 12x + 32 # 213x + 42 # 3 129. 2x12x + 52 + x2 # 2 128. 512x + 12 2 + 15x - 62 # 212x + 12 # 2 130. 3x2 18x - 32 + x3 # 8 131. 21x + 32 1x - 22 3 + 1x + 32 2 # 31x - 22 2 132. 41x + 52 3 1x - 12 2 + 1x + 52 4 # 21x - 12 135. 213x - 52 # 312x + 12 3 + 13x - 52 2 # 312x + 12 2 # 2 136. 314x + 52 2 # 415x + 12 2 + 14x + 52 3 # 215x + 12 # 5 133. 14x - 32 2 + x # 214x - 32 # 4 137. Show that x2 + 4 is prime. 134. 3x2 13x + 42 2 + x3 # 213x + 42 # 3 138. Show that x2 + x + 1 is prime. Explaining Concepts: Discussion and Writing 139. Make up a polynomial that factors into a perfect square. 140. Explain to a fellow student what you look for first when presented with a factoring problem. What do you do next? R.6 Synthetic Division OBJECTIVE 1 Divide Polynomials Using Synthetic Division (p. 58) 1 Divide Polynomials Using Synthetic Division To find the quotient as well as the remainder when a polynomial of degree 1 or higher is divided by x - c, a shortened version of long division, called synthetic division, makes the task simpler. SECTION R.6 Synthetic Division 59 To see how synthetic division works, first consider long division for dividing the polynomial 2x3 - x2 + 3 by x - 3. d Quotient 2x2 + 5x + 15 3 2 x - 3 ) 2x - x + 3 2x3 - 6x2 5x2 5x2 - 15x 15x + 3 15x - 45 48 d Remainder Check: 1Divisor2 # 1Quotient2 + Remainder = 1x - 32 12x2 + 5x + 152 + 48 = 2x3 + 5x2 + 15x - 6x2 - 15x - 45 + 48 = 2x3 - x2 + 3 The process of synthetic division arises from rewriting the long division in a more compact form, using simpler notation. For example, in the long division above, the terms in blue are not really necessary because they are identical to the terms directly above them. With these terms removed, we have 2x2 + 5x + 15 x - 3 ) 2x3 - x2 + 3 2 - 6x 5x2 - 15x 15x - 45 48 Most of the x’s that appear in this process can also be removed, provided that we are careful about positioning each coefficient. In this regard, we will need to use 0 as the coefficient of x in the dividend, because that power of x is missing. Now we have 2x2 + 5x + 15 x - 3) 2 - 1 0 - 6 5 - 15 15 3 - 45 48 We can make this display more compact by moving the lines up until the numbers in blue align horizontally. 2x2 + 5x + 15 0 3 x - 3) 2 - 1 - 6 - 15 - 45 ~ 5 15 48 Row 1 Row 2 Row 3 Row 4 Because the leading coefficient of the divisor is always 1, the leading coefficient of the dividend will also be the leading coefficient of the quotient. So we place the leading coefficient of the quotient, 2, in the circled position. Now, the first three numbers in row 4 are precisely the coefficients of the quotient, and the last number CHAPTER R Review in row 4 is the remainder. Since row 1 is not really needed, we can compress the process to three rows, where the bottom row contains both the coefficients of the quotient and the remainder. x - 3) 2 - 1 0 3 Row 1 - 6 - 15 - 45 Row 2 (subtract) 5 15 48 Row 3 2 3 45 48 S3 0 15 15 * S3 * 2 - 1 6 5 * x - 3) 2 S3 Recall that the entries in row 3 are obtained by subtracting the entries in row 2 from those in row 1. Rather than subtracting the entries in row 2, we can change the sign of each entry and add. With this modification, our display will look like this: Row 1 Row 2 (add) Row 3 Notice that the entries in row 2 are three times the prior entries in row 3. Our last modification to the display replaces the x - 3 by 3. The entries in row 3 give the quotient and the remainder, as shown next. 3) 2 0 15 15 -1 6 5 3 45 48 Row 1 Row 2 (add) Row 3 c e 2 Quotient Remainder c e 2 2x + 5x + 15 48 Let’s go through an example step by step. EX AMPLE 1 Using Synthetic Division to Find the Quotient and Remainder Use synthetic division to find the quotient and remainder when x3 - 4x2 - 5 is divided by x - 3 STEP 1: Write the dividend in descending powers of x. Then copy the coefficients, remembering to insert a 0 for any missing powers of x. 1 -4 0 -5 Row 1 STEP 2: Insert the usual division symbol. In synthetic division, the divisor is of the form x - c, and c is the number placed to the left of the division symbol. Here, since the divisor is x - 3, insert 3 to the left of the division symbol. 3) 1 -4 0 -5 Row 1 STEP 3: Bring the 1 down two rows, and enter it in row 3. 3) 1 - 4 0 - 5 T 1 STEP 4: Multiply the latest entry in row 3 by column over to the right. 3) 1 -4 0 3 Row 1 Row 2 Row 3 3, and place the result in row 2, one - 5 Row 1 Row 2 Row 3 * S3 1 STEP 5: Add the entry in row 2 to the entry above it in row 1, and enter the sum in row 3. 3 ) 1 - 4 0 - 5 Row 1 Row 2 3 Row 3 1 -1 S3 Solution * 60 SECTION R.6 Synthetic Division 61 STEP 6: Repeat Steps 4 and 5 until no more entries are available in row 1. 3) 1 S3 * S3 * * S3 -4 0 - 5 Row 1 3 - 3 - 9 Row 2 1 - 1 - 3 - 14 Row 3 STEP 7: The final entry in row 3, the - 14, is the remainder; the other entries in row 3, the 1, - 1, and - 3, are the coefficients (in descending order) of a polynomial whose degree is 1 less than that of the dividend. This is the quotient. That is, Quotient = x2 - x - 3 Remainder = - 14 Check: 1Divisor2 1Quotient2 + Remainder = 1x - 32 1x2 - x - 32 + 1 - 142 = 1x3 - x2 - 3x - 3x2 + 3x + 92 + 1 - 142 r = x3 - 4x2 - 5 = Dividend Let’s do an example in which all seven steps are combined. EXAMPL E 2 Using Synthetic Division to Verify a Factor Use synthetic division to show that x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3 Solution The divisor is x + 3 = x - ( - 3), so place - 3 to the left of the division symbol. Then the row 3 entries will be multiplied by - 3, entered in row 2, and added to row 1. - 3) 2 2 5 -6 -1 -2 3 1 2 -3 -1 -2 3 1 3 -3 0 Row 1 Row 2 Row 3 Because the remainder is 0, we have 1Divisor2 1Quotient2 + Remainder = 1x + 32 12x4 - x3 + x2 - x + 12 = 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3 r As we see, x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3. As Example 2 illustrates, the remainder after division gives information about whether the divisor is, or is not, a factor. We shall have more to say about this in Chapter 5. Now Work PROBLEMS 9 AND 19 R.6 Assess Your Understanding Concepts and Vocabulary 1. To check division, the expression that is being divided, the dividend, should equal the product of the the . and the 2. To divide 2x3 - 5x + 1 by x + 3 using synthetic division, the first step is to write . ) 3. Choose the division problem that cannot be done using synthetic division. (b) x4 - 3 is divided by x + 1 (a) 2x3 - 4x2 + 6x - 8 is divided by x - 8 5 2 (c) x + 3x - 9x + 2 is divided by x + 10 (d) x4 - 5x3 + 3x2 - 9x + 13 is divided by 4. Choose the correct conclusion based on the following synthetic division: - 5 ) 2 3 - 38 - 15 x2 + 5 - 10 35 15 2 -7 - 3 0 (b) x - 5 is a factor of 2x3 + 3x2 - 38x - 15 (a) x + 5 is a factor of 2x3 + 3x2 - 38x - 15 (c) x + 5 is not a factor of 2x3 + 3x2 - 38x - 15 (d) x - 5 is not a factor of 2x3 + 3x2 - 38x - 15 5. True or False In using synthetic division, the divisor is always a polynomial of degree 1, whose leading coefficient is 1. 6. True or False - 2 ) 5 3 2 1 5x3 + 3x2 + 2x + 1 - 31 = 5x2 - 7x + 16 + . - 10 14 - 32 means x + 2 x + 2 5 - 7 16 - 31 plus 62 CHAPTER R Review Skill Building In Problems 7–18, use synthetic division to find the quotient and remainder when: 7. x3 - x2 + 2x + 4 is divided by x - 2 8. x3 + 2x2 - 3x + 1 is divided by x + 1 9. 3x3 + 2x2 - x + 3 is divided by x - 3 10. - 4x3 + 2x2 - x + 1 is divided by x + 2 11. x5 - 4x3 + x is divided by x + 3 12. x4 + x2 + 2 is divided by x - 2 13. 4x6 - 3x4 + x2 + 5 is divided by x - 1 14. x5 + 5x3 - 10 is divided by x + 1 15. 0.1x3 + 0.2x is divided by x + 1.1 16. 0.1x2 - 0.2 is divided by x + 2.1 17. x5 - 1 is divided by x - 1 18. x5 + 1 is divided by x + 1 In Problems 19–28, use synthetic division to determine whether x - c is a factor of the given polynomial. 19. 4x3 - 3x2 - 8x + 4; x - 2 20. - 4x3 + 5x2 + 8; x + 3 21. 3x4 - 6x3 - 5x + 10; x - 2 22. 4x4 - 15x2 - 4; x - 2 23. 3x6 + 82x3 + 27; x + 3 24. 2x6 - 18x4 + x2 - 9; x + 3 25. 4x6 - 64x4 + x2 - 15; x + 4 1 27. 2x4 - x3 + 2x - 1; x 2 26. x6 - 16x4 + x2 - 16; x + 4 1 28. 3x4 + x3 - 3x + 1; x + 3 Applications and Extensions 29. Find the sum of a, b, c, and d if x3 - 2x2 + 3x + 5 d = ax2 + bx + c + x + 2 x + 2 Explaining Concepts: Discussion and Writing 30. When dividing a polynomial by x - c, do you prefer to use long division or synthetic division? Does the value of c make a difference to you in choosing? Give reasons. R.7 Rational Expressions OBJECTIVES 1 Reduce a Rational Expression to Lowest Terms (p. 62) 2 Multiply and Divide Rational Expressions (p. 63) 3 Add and Subtract Rational Expressions (p. 64) 4 Use the Least Common Multiple Method (p. 66) 5 Simplify Complex Rational Expressions (p. 68) 1 Reduce a Rational Expression to Lowest Terms If we form the quotient of two polynomials, the result is called a rational expression. Some examples of rational expressions are (a) x3 + 1 x (b) 3x2 + x - 2 x2 + 5 (c) x 2 x - 1 (d) xy2 1x - y2 2 Expressions (a), (b), and (c) are rational expressions in one variable, x, whereas (d) is a rational expression in two variables, x and y. Rational expressions are described in the same manner as rational numbers. In expression (a), the polynomial x3 + 1 is the numerator, and x is the denominator. When the numerator and denominator of a rational expression contain no common factors (except 1 and - 1), we say that the rational expression is reduced to lowest terms, or simplified. The polynomial in the denominator of a rational expression cannot be equal x3 + 1 to 0 because division by 0 is not defined. For example, for the expression ,x x cannot take on the value 0. The domain of the variable x is 5 x 0 x ≠ 06 . SECTION R.7 Rational Expressions 63 A rational expression is reduced to lowest terms by factoring the numerator and the denominator completely and canceling any common factors using the Cancellation Property: ac a = bc b EXAMPL E 1 (1) Reducing a Rational Expression to Lowest Terms Reduce to lowest terms: Solution if b ≠ 0, c ≠ 0 x2 + 4x + 4 x2 + 3x + 2 Begin by factoring the numerator and the denominator. x2 + 4x + 4 = 1x + 22 1x + 22 x2 + 3x + 2 = 1x + 22 1x + 12 WARNING Apply the Cancellation Property only to rational expressions written in factored form. Be sure to cancel only common factors, not common terms! ■ EXAMPL E 2 Since a common factor, x + 2, appears, the original expression is not in lowest terms. To reduce it to lowest terms, use the Cancellation Property: 1x + 22 1x + 22 x + 2 x2 + 4x + 4 = = 2 1x + 22 1x + 12 x + 1 x + 3x + 2 x ≠ - 2, - 1 r Reducing Rational Expressions to Lowest Terms Reduce each rational expression to lowest terms. (a) Solution (a) (b) x3 - 8 x3 - 2x2 (b) 8 - 2x x - x - 12 2 1x - 22 1x2 + 2x + 42 x2 + 2x + 4 x3 - 8 = = x3 - 2x2 x2 1x - 22 x2 x ≠ 0, 2 21 - 12 1x - 42 214 - x2 8 - 2x -2 = = = 1x 42 1x + 32 1x 42 1x + 32 x + 3 x - x - 12 2 Now Work PROBLEM x ≠ - 3, 4 r 7 2 Multiply and Divide Rational Expressions The rules for multiplying and dividing rational expressions are the same as the rules a c for multiplying and dividing rational numbers. If and , b ≠ 0, d ≠ 0, are two b d rational expressions, then a#c ac = b d bd a b a d ad = # = c b c bc d if b ≠ 0, d ≠ 0 if b ≠ 0, c ≠ 0, d ≠ 0 (2) (3) In using equations (2) and (3) with rational expressions, be sure first to factor each polynomial completely so that common factors can be canceled. Leave your answer in factored form. 64 CHAPTER R Review EX AMPLE 3 Multiplying and Dividing Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. x + 3 2 2 x - 2x + 1 # 4x + 4 x2 - 4 (b) 2 (a) 3 2 x + x x + x - 2 x - x - 12 x3 - 8 Solution (a) 1x - 12 2 41x2 + 12 x2 - 2x + 1 # 4x2 + 4 # = x3 + x x2 + x - 2 x1x2 + 12 1x + 22 1x - 12 1x - 12 2 142 1x2 + 12 = x 1x2 + 12 1x + 22 1x - 12 41x - 12 = x ≠ - 2, 0, 1 x1x + 22 x + 3 x + 3 # x3 - 8 x2 - 4 = 2 (b) 2 x - 4 x2 - x - 12 x - x - 12 x3 - 8 = 2 x + 3 # 1x - 22 1x + 2x + 42 1x - 22 1x + 22 1x - 42 1x + 32 = 1x + 32 1x - 22 1x2 + 2x + 42 1x - 22 1x + 22 1x - 42 1x + 32 = x2 + 2x + 4 1x + 22 1x - 42 Now Work PROBLEMS 19 x ≠ - 3, - 2, 2, 4 AND r 27 3 Add and Subtract Rational Expressions In Words To add (or subtract) two rational expressions with the same denominator, keep the common denominator and add (or subtract) the numerators. The rules for adding and subtracting rational expressions are the same as the rules for adding and subtracting rational numbers. If the denominators of two rational expressions to be added (or subtracted) are equal, then add (or subtract) the numerators and keep the common denominator. If a c and are two rational expressions, then b b c a + c a + = b b b EX AMPLE 4 c a - c a = b b b if b ≠ 0 (4) Adding and Subtracting Rational Expressions with Equal Denominators Perform the indicated operation and simplify the result. Leave your answer in factored form. (a) Solution (a) 2x2 - 4 x + 3 + 2x + 5 2x + 5 x ≠ - 5 2 (b) x 3x + 2 x - 3 x - 3 12x2 - 42 + 1x + 32 2x2 - 4 x + 3 + = 2x + 5 2x + 5 2x + 5 2 12x - 12 1x + 12 2x + x - 1 = = 2x + 5 2x + 5 x ≠ 3 SECTION R.7 Rational Expressions (b) EXAMPL E 5 Solution x - 13x + 22 x 3x + 2 x - 3x - 2 = = x - 3 x - 3 x - 3 x - 3 21x + 12 - 2x - 2 = = x - 3 x - 3 65 r Adding Rational Expressions Whose Denominators Are Additive Inverses of Each Other Perform the indicated operation and simplify the result. Leave your answer in factored form. 2x 5 + x ≠ 3 x - 3 3 - x Notice that the denominators of the two rational expressions are different. However, the denominator of the second expression is the additive inverse of the denominator of the first. That is, 3 - x = - x + 3 = - 1 # 1x - 32 = - 1x - 32 Then 2x 5 2x 5 2x -5 + = + = + x - 3 3 - x x - 3 - 1x - 32 x - 3 x - 3 c a c -a 3 - x = - (x - 3) = -b b 2x + 1 - 52 2x - 5 = = x - 3 x - 3 Now Work PROBLEMS 39 AND r 45 If the denominators of two rational expressions to be added or subtracted are not equal, we can use the general formulas for adding and subtracting rational expressions. a c a#d b#c ad + bc + = # + # = b d b d b d bd # # a c a d b c ad - bc = # - # = b d b d b d bd EXAMPL E 6 if b ≠ 0, d ≠ 0 (5a) if b ≠ 0, d ≠ 0 (5b) Adding and Subtracting Rational Expressions with Unequal Denominators Perform the indicated operation and simplify the result. Leave your answer in factored form. x x x (a) x (a) Solution + + 3 x x2 1 + x ≠ - 4, 2 (b) 2 x 4 x - 2 x - 4 3 x x - 3#x - 2 x + 4# x + = + 4 x - 2 x + 4 x - 2 x + 4 x - 2 æ (5a) x ≠ - 2, 0, 2 = 1x - 32 1x - 22 + 1x + 42 1x2 1x + 42 1x - 22 = 2x2 - x + 6 x2 - 5x + 6 + x2 + 4x = 1x + 42 1x - 22 1x + 42 1x - 22 66 CHAPTER R Review (b) x2 1x2 - 1x2 - 42 112 x2 x2 # x x2 - 4 # 1 1 = = x x2 - 4 x2 - 4 x x2 - 4 x 1x2 - 42 1x2 æ (5b) = x3 - x2 + 4 1x - 22 1x + 22 1x2 Now Work PROBLEM r 49 4 Use the Least Common Multiple Method If the denominators of two rational expressions to be added (or subtracted) have common factors, we usually do not use the general rules given by equations (5a) and (5b). Just as with fractions, we apply the least common multiple (LCM) method. The LCM method uses the polynomial of least degree that has each denominator polynomial as a factor. The LCM Method for Adding or Subtracting Rational Expressions The Least Common Multiple (LCM) Method requires four steps: STEP 1: STEP 2: STEP 3: STEP 4: Factor completely the polynomial in the denominator of each rational expression. The LCM of the denominators is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. Write each rational expression using the LCM as the common denominator. Add or subtract the rational expressions using equation (4). We begin with an example that requires only Steps 1 and 2. EX AMPLE 7 Finding the Least Common Multiple Find the least common multiple of the following pair of polynomials: x1x - 12 2 1x + 12 Solution and 41x - 12 1x + 12 3 STEP 1: The polynomials are already factored completely as x1x - 12 2 1x + 12 and 41x - 12 1x + 12 3 STEP 2: Start by writing the factors of the left-hand polynomial. (Or you could start with the one on the right.) x1x - 12 2 1x + 12 Now look at the right-hand polynomial. Its first factor, 4, does not appear in our list, so we insert it. 4x1x - 12 2 1x + 12 The next factor, x - 1, is already in our list, so no change is necessary. The final factor is 1x + 12 3. Since our list has x + 1 to the first power only, we replace x + 1 in the list by 1x + 12 3. The LCM is 4x1x - 12 2 1x + 12 3 SECTION R.7 Rational Expressions 67 Notice that the LCM is, in fact, the polynomial of least degree that contains x1x - 12 2 1x + 12 and 41x - 12 1x + 12 3 as factors. r Now Work PROBLEM 55 Using the Least Common Multiple to Add Rational Expressions EXAMPL E 8 Perform the indicated operation and simplify the result. Leave your answer in factored form. x 2x - 3 + 2 x2 + 3x + 2 x - 1 Solution x ≠ - 2, - 1, 1 STEP 1: Factor completely the polynomials in the denominators. x2 + 3x + 2 = 1x + 22 1x + 12 x2 - 1 = 1x - 12 1x + 12 STEP 2: The LCM is 1x + 22 1x + 12 1x - 12 . Do you see why? STEP 3: Write each rational expression using the LCM as the denominator. x1x - 12 x x x #x - 1 = = = 1x + 22 1x + 12 1x + 22 1x + 12 x - 1 1x + 22 1x + 12 1x - 12 x + 3x + 2 2 æ Multiply numerator and denominator by x - 1 to get the LCM in the denominator. 2x - 3 2x - 3 2x - 3 # x + 2 = 12x - 32 1x + 22 = = 2 1x - 12 1x + 12 1x - 12 1x + 12 x + 2 1x - 12 1x + 12 1x + 22 x - 1 æ Multiply numerator and denominator by x + 2 to get the LCM in the denominator. STEP 4: Now add by using equation (4). 12x - 32 1x + 22 x1x - 12 x 2x - 3 + + 2 = 1x + 22 1x + 12 1x - 12 1x + 22 1x + 12 1x - 12 x2 + 3x + 2 x - 1 EXAMPL E 9 = 1x2 - x2 + 12x2 + x - 62 1x + 22 1x + 12 1x - 12 = 31x2 - 22 3x2 - 6 = 1x + 22 1x + 12 1x - 12 1x + 22 1x + 12 1x - 12 r Using the Least Common Multiple to Subtract Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. 3 x + 4 - 2 x2 + x x + 2x + 1 Solution x ≠ - 1, 0 STEP 1: Factor completely the polynomials in the denominators. x2 + x = x1x + 12 x2 + 2x + 1 = 1x + 12 2 STEP 2: The LCM is x1x + 12 2. 68 CHAPTER R Review STEP 3: Write each rational expression using the LCM as the denominator. 3 3 3 # x + 1 = 31x + 122 = = x1x + 12 x1x + 12 x + 1 x + x x1x + 12 2 x1x + 42 x + 4 x + 4 x + 4 #x = = = 2 2 x x + 2x + 1 1x + 12 1x + 12 x1x + 12 2 2 STEP 4: Subtract, using equation (4). 31x + 12 x1x + 42 3 x + 4 - 2 = 2 x + x x + 2x + 1 x1x + 12 x1x + 12 2 2 = Now Work PROBLEM 31x + 12 - x1x + 42 x1x + 12 2 = 3x + 3 - x2 - 4x x1x + 12 2 = - x2 - x + 3 x1x + 12 2 r 65 5 Simplify Complex Rational Expressions When sums and/or differences of rational expressions appear as the numerator and/or denominator of a quotient, the quotient is called a complex rational expression.* For example, 1 x 1 1 x 1 + and x2 - 3 x2 - 4 x - 3 - 1 x + 2 are complex rational expressions. To simplify a complex rational expression means to write it as a rational expression reduced to lowest terms. This can be accomplished in either of two ways. Simplifying a Complex Rational Expression OPTION 1: Treat the numerator and denominator of the complex rational expression separately, performing whatever operations are indicated and simplifying the results. Follow this by simplifying the resulting rational expression. OPTION 2: Find the LCM of the denominators of all rational expressions that appear in the complex rational expression. Multiply the numerator and denominator of the complex rational expression by the LCM and simplify the result. We use both options in the next example. By carefully studying each option, you can discover situations in which one may be easier to use than the other. EX A MPL E 1 0 Simplifying a Complex Rational Expression 1 3 + x 2 Simplify: x + 3 4 x ≠ - 3, 0 * Some texts use the term complex fraction. SECTION R.7 Rational Expressions Solution 69 Option 1: First, we perform the indicated operation in the numerator, and then we divide. 1 3 1#x + 2#3 x + 6 + x 2 2#x 2x x + 6# 4 = = = x + 3 x + 3 x + 3 2x x + 3 æ 4 æ 4 4 Rule for adding quotients = Rule for dividing quotients 1x + 62 # 4 2 # 2 # 1x + 62 21x + 62 = = 2 # x # 1x + 32 2 # x # 1x + 32 x1x + 32 æ Rule for multiplying quotients Option 2: The rational expressions that appear in the complex rational expression are 1 , 2 3 , x x + 3 4 The LCM of their denominators is 4x. We multiply the numerator and denominator of the complex rational expression by 4x and then simplify. 1 3 3 1 3 1 4x # a + b + 4x # + 4x # x x x 2 2 2 = = # 4x 1x + 32 x + 3 x + 3 4x # a b 4 4 4 æ Multiply the æ Use the Distributive Property numerator and in the numerator. denominator by 4x. 1 3 + 4x # 21x + 62 x 2x + 12 2 = = # 4 x 1x + 32 x1x + 32 x1x + 32 æ æ 4 2 # 2x # = Simplify EX AM PL E 11 r Simplifying a Complex Rational Expression x2 + 2 x - 4 Simplify: 2x - 2 - 1 x Solution Factor x ≠ 0, 2, 4 We will use Option 1. 21x - 42 x2 x2 + 2x - 8 x2 + 2 + x - 4 x - 4 x - 4 x - 4 = = 2x - 2 2x - 2 x 2x - 2 - x - 1 x x x x 1x + 42 1x - 22 1x + 42 1x - 22 x - 4 = = x - 2 x - 4 x = Now Work 1x + 42 # x x - 4 PROBLEM 75 # x x - 2 r 70 CHAPTER R Review Application Solving an Application in Electricity EX A MPL E 12 An electrical circuit contains two resistors connected in parallel, as shown in Figure 28. If these two resistors provide resistance of R1 and R2 ohms, respectively, their combined resistance R is given by the formula R1 R = R2 Figure 28 1 1 1 + R1 R2 Express R as a rational expression; that is, simplify the right-hand side of this formula. Evaluate the rational expression if R1 = 6 ohms and R2 = 10 ohms. Solution 1 We will use Option 2. If we consider 1 as the fraction , the rational expressions in 1 the complex rational expression are 1 1 1 , , 1 R1 R2 The LCM of the denominators is R1 R2 . We multiply the numerator and denominator of the complex rational expression by R1 R2 and simplify. 1 1 1 + R1 R2 = 1 # R1 R2 a 1 1 # + b R1 R2 R1 R2 = R1 R2 1 # 1 # RR + RR R1 1 2 R2 1 2 = R1 R2 R2 + R1 So, R = R1 R2 R2 + R1 If R1 = 6 and R2 = 10, then R = 6 # 10 60 15 = = 10 + 6 16 4 ohms r R.7 Assess Your Understanding Concepts and Vocabulary 1. When the numerator and denominator of a rational expression contain no common factors (except 1 and - 1), the rational expression is in . 2. LCM is an abbreviation for . 3. Choose the statement that is not true. Assume b ≠ 0, c ≠ 0, and d ≠ 0 as necessary. a a c a + c ac (a) = (b) + = bc b b b b a a b ac c ad - bc (c) = = (d) b d bd c bd d 4. Choose the rational expression that simplifies to - 1. a - b a - b (a) (b) b - a a - b a + b b - a (c) (d) a - b b + a 2x3 - 4x is reduced 5. True or False The rational expression x - 2 to lowest terms. 6. True or False The LCM of 2x3 + 6x2 and 6x4 + 4x3 is 4x3(x + 1). Skill Building In Problems 7–18, reduce each rational expression to lowest terms. 7. 3x + 9 x2 - 9 8. 4x2 + 8x 12x + 24 9. 11. 24x2 12x2 - 6x 12. x2 + 4x + 4 x2 - 4 13. 15. x2 + 4x - 5 x2 - 2x + 1 16. x - x2 x2 + x - 2 17. x2 - 2x 3x - 6 y2 - 25 2 2y - 8y - 10 x2 + 5x - 14 2 - x 10. 14. 18. 15x2 + 24x 3x2 3y2 - y - 2 3y2 + 5y + 2 2x2 + 5x - 3 1 - 2x SECTION R.7 Rational Expressions 71 In Problems 19–36, perform the indicated operation and simplify the result. Leave your answer in factored form. 19. 3x + 6 # x 5x2 x2 - 4 23. 12 4x - 8 # - 3x 12 - 6x 20. 3 # x2 2x 6x + 10 24. 21. 4x2 # x3 - 64 2x x2 - 16 2 6x - 27 # 5x 4x - 18 25. 6x x - 4 27. 3x - 9 2x + 4 x2 + x - 6 # x2 - 25 26. 2 x + 4x - 5 x2 + 2x - 15 2 28. 30. x2 + 7x + 12 x2 - 7x + 12 33. 2 x + x - 12 x2 - x - 12 x2 + 7x + 6 x2 + x - 6 34. 2 x + 5x - 6 x2 + 5x + 6 12 # x3 + 1 x2 + x 4x - 2 x2 - 3x - 10 # x2 + 4x - 21 x2 + 2x - 35 x2 + 9x + 14 12x 5x + 20 4x2 x - 16 2 x - 2 4x 8x x2 - 1 29. 10x x + 1 22. x2 - 4x + 4 12x 4 - x 4 + x 31. 4x 2 x - 16 32. 2x2 - x - 28 3x2 - x - 2 35. 4x2 + 16x + 7 3x2 + 11x + 6 9x2 + 3x - 2 12x2 + 5x - 2 36. 9x2 - 6x + 1 8x2 - 10x - 3 3 + x 3 - x x2 - 9 9x3 In Problems 37–54, perform the indicated operation and simplify the result. Leave your answer in factored form. 37. 5 x + 2 2 38. 6 3 x x 39. x2 4 2x - 3 2x - 3 40. 9 3x2 2x - 1 2x - 1 41. x + 1 2x - 3 + x - 3 x - 3 42. 2x - 5 x + 4 + 3x + 2 3x + 2 43. 3x + 5 2x - 4 2x - 1 2x - 1 44. 5x - 4 x + 1 3x + 4 3x + 4 45. x 4 + x - 2 2 - x 46. 6 x x - 1 1 - x 47. 4 2 x - 1 x + 2 48. 5 2 x + 5 x - 5 49. 2x - 3 x + x + 1 x - 1 50. 3x 2x + x - 4 x + 3 51. x - 3 x + 4 x + 2 x - 2 52. 2x - 3 2x + 1 x - 1 x + 1 53. 1 x + x x2 - 4 54. x - 1 x + 2 x + 1 x3 In Problems 55–62, find the LCM of the given polynomials. 55. x2 - 4, x2 - x - 2 56. x2 - x - 12, x2 - 8x + 16 57. x3 - x, x2 - x 58. 3x2 - 27, 2x2 - x - 15 59. 4x3 - 4x2 + x, 2x3 - x2, x3 60. x - 3, x2 + 3x, x3 - 9x 61. x3 - x, x3 - 2x2 + x, x3 - 1 62. x2 + 4x + 4, x3 + 2x2, 1x + 22 3 In Problems 63–74, perform the indicated operations and simplify the result. Leave your answer in factored form. 63. x x - 2 x2 - 7x + 6 x - 2x - 24 64. 66. x - 4 3x - 2 x - 1 x - 2x + 1 67. 1x - 12 1x + 12 69. 2x + 3 x + 4 - 2 x2 - x - 2 x + 2x - 8 70. 72. x + 1 x 2 + - 3 x 1x - 12 2 x - x2 73. x x + 1 - 2 x - 3 x + 5x - 24 65. 2 4x - 2 x2 - 4 x + x - 6 68. 6 2 1x + 22 2 1x - 12 1x + 22 1x - 12 2 2x - 3 x - 2 x2 + 8x + 7 1x + 12 2 71. 3 1 2 + 3 - 2 x x + x x - x2 1 1 1 a - b h x + h x 74. 1 1 1 J - 2R h 1x + h2 2 x 3 2 + 2 1x - 12 1x + 12 2 72 CHAPTER R Review In Problems 75–86, perform the indicated operations and simplify the result. Leave your answer in factored form. x x + 1 78. x - 1 2 x 1 x 75. 1 1 x 1 x2 76. 1 3 - 2 x x + 1 x 77. x - 1 3 + x + 1 x - 3 x + 4 x - 2 x + 1 79. x + 1 x - 2 x x + 1 x - 2 80. x + 3 x - 2 x - 1 + x + 2 x + 1 81. x 2x - 3 x + 1 x 1 + 83. 1 - 4 + 1 1 - 1 84. 1 - 1 x 1 - 2 - 1 - 85. 1 1 - x 82. 21x - 12 - 1 + 3 31x - 12 -1 86. + 2 2x + 5 x x x - 3 1x + 12 2 x2 x - 3 x + 3 41x + 22 - 1 - 3 31x + 22 - 1 - 1 In Problems 87–94, expressions that occur in calculus are given. Reduce each expression to lowest terms. 87. 90. 93. 12x + 32 # 3 - 13x - 52 # 2 13x - 52 2 88. x # 2x - 1x2 - 42 # 1 1x2 - 42 91. 2 14x + 12 # 5 - 15x - 22 # 4 15x - 22 2 13x + 12 # 2x - x2 # 3 13x + 12 2 1x2 + 12 # 3 - 13x + 42 # 2x 1x2 + 12 89. 92. 1x2 + 12 2 12x - 52 # 3x2 - x3 # 2 12x - 52 2 1x2 + 92 # 2 - 12x - 52 # 2x 94. 2 x # 2x - 1x2 + 12 # 1 1x2 + 92 2 Applications and Extensions 96. Electrical Circuits An electrical circuit contains three resistors connected in parallel. If these three resistors provide resistance of R1 , R2 , and R3 ohms, respectively, their combined resistance R is given by the formula 95. The Lensmaker’s Equation The focal length f of a lens with index of refraction n is 1 1 1 = 1n - 12 J + R f R1 R2 1 1 1 1 = + + R R1 R2 R3 where R1 and R2 are the radii of curvature of the front and back surfaces of the lens. Express f as a rational expression. Evaluate the rational expression for n = 1.5, R1 = 0.1 meter, and R2 = 0.2 meter. Express R as a rational expression. Evaluate R for R1 = 5 ohms, R2 = 4 ohms, and R3 = 10 ohms. Explaining Concepts: Discussion and Writing 97. The following expressions are called continued fractions: 1 + 1 , 1 + x 1 1 + 1 x 1 , 1 + 1 + 1 1 + 1 , 1 + 1 x ,... 1 1 + 1 + 1 1 + 1 x Each simplifies to an expression of the form ax + b bx + c Trace the successive values of a, b, and c as you “continue” the fraction. Can you discover the patterns that these values follow? Go to the library and research Fibonacci numbers. Write a report on your findings. 98. Explain to a fellow student when you would use the LCM method to add two rational expressions. Give two examples of adding two rational expressions: one in which you use the LCM and the other in which you do not. 99. Which of the two options given in the text for simplifying complex rational expressions do you prefer? Write a brief paragraph stating the reasons for your choice. SECTION R.8 nth Roots; Rational Exponents 73 R.8 nth Roots; Rational Exponents PREPARING FOR THIS SECTION Before getting started, review the following: r Exponents, Square Roots (Section R.2, pp. 21–24) Now Work the ‘Are You Prepared?’ problems on page 78. OBJECTIVES 1 Work with nth Roots (p. 73) 2 Simplify Radicals (p. 74) 3 Rationalize Denominators (p. 75) 4 Simplify Expressions with Rational Exponents (p. 76) 1 Work with nth Roots DEFINITION The principal nth root of a real number a, n Ú 2 an integer, symbolized by n 2a, is defined as follows: n 2a = b means a = bn where a Ú 0 and b Ú 0 if n is even and a, b are any real numbers if n is odd. n In Words n The symbol 2a means “give me the number that, when raised to the power n, equals a.” EXAMPL E 1 Notice that if a is negative and n is even, then 2a is not defined. When it is defined, the principal nth root of a number is unique. n The symbol 2a for the principal nth root of a is called a radical; the integer n is called the index, and a is called the radicand. If the index of a radical is 2, we call 2 2 a the square root of a and omit the index 2 by simply writing 1a. If the index is 3 3, we call 2 a the cube root of a. Simplifying Principal nth Roots 3 3 3 (a) 2 8 = 2 2 = 2 (c) 3 3 (b) 2 - 64 = 2 1 - 42 3 = - 4 6 (d) 2 1 - 22 6 = 0 - 2 0 = 2 1 1 4 1 = 4 a b = A 16 2 B 2 4 r These are examples of perfect roots, since each simplifies to a rational number. Notice the absolute value in Example 1(d). If n is even, then the principal nth root must be nonnegative. In general, if n Ú 2 is an integer and a is a real number, we have n 2a n = a 2a n = 0 a 0 n Now Work PROBLEM if n Ú 3 is odd (1a) if n Ú 2 is even (1b) 11 Radicals provide a way of representing many irrational real numbers. For example, it can be shown that there is no rational number whose square is 2. Using radicals, we can say that 12 is the positive number whose square is 2. 74 CHAPTER R Review EX AMPLE 2 Using a Calculator to Approximate Roots 5 Use a calculator to approximate 2 16. Solution Figure 29 shows the result using a TI-84 Plus C graphing calculator. Now Work r 113 PROBLEM 2 Simplify Radicals Let n Ú 2 and m Ú 2 denote integers, and let a and b represent real numbers. Assuming that all radicals are defined, we have the following properties: Figure 29 Properties of Radicals n n n 2ab = 2a 2b (2a) n 2a n a = n Ab 2b n b ≠ 0 n 2am = ( 2a2 m (2b) (2c) When used in reference to radicals, the direction to “simplify” will mean to remove from the radicals any perfect roots that occur as factors. EX AMPLE 3 Simplifying Radicals (a) 232 = 216 # 2 = 216 # 22 = 422 c c (2a) Factor out 16, a perfect square. 3 3 3 3 3 3# 3 3 (b) 2 16 = 2 8#2 = 2 8# 2 2 = 2 2 22 = 22 2 c c Factor out 8, a perfect cube. (2a) 3 3 3 (c) 2 - 16x4 = 2 - 8 # 2 # x3 # x = 2 1 - 8x3 2 12x2 c c Group perfect Factor perfect cubes inside radical. cubes. 3 3 3 3 = 2 1 - 2x2 3 # 2x = 2 1 - 2x2 3 # 2 2x = - 2x2 2x c (2a) (d) 16x5 24 x4 x 2x 4 # 2x 4 # 4 2x 4 4 4 = 4 = a b x = a b 2x = ` ` 2 x 4 3 B 81 B 3 B 3 B 3 4 Now Work PROBLEMS 15 AND 21 r Two or more radicals can be combined, provided that they have the same index and the same radicand. Such radicals are called like radicals. EX AMPLE 4 Combining Like Radicals (a) - 8212 + 23 = - 824 # 3 + 23 = - 8 # 24 23 + 23 = - 1623 + 23 = - 1523 SECTION R.8 nth Roots; Rational Exponents 75 3 3 3 3 3 3 3 3 3 (b) 2 8x4 + 2 - x + 42 27x = 2 2xx + 2 - 1 # x + 42 3 x 3 3 3 3 3 3# 3 = 2 12x2 3 # 2 x + 2 -1 # 2 x + 42 3 2x 3 3 3 = 2x2 x - 1# 2 x + 122 x 3 = 12x + 112 2 x Now Work PROBLEM r 39 3 Rationalize Denominators When radicals occur in quotients, it is customary to rewrite the quotient so that the new denominator contains no radicals. This process is referred to as rationalizing the denominator. The idea is to multiply by an appropriate expression so that the new denominator contains no radicals. For example: If a Denominator Contains the Factor Multiply by To Obtain a Denominator Free of Radicals 23 23 23 + 1 23 - 1 22 - 3 22 + 3 25 - 23 25 + 23 1 252 - 1 232 2 = 5 - 3 = 2 3 2 4 3 2 2 3 3 3 2 4# 2 2 = 2 8 = 2 1 232 = 3 2 1 232 - 12 = 3 - 1 = 2 2 1 222 - 32 = 2 - 9 = - 7 2 2 In rationalizing the denominator of a quotient, be sure to multiply both the numerator and the denominator by the expression. EXAMPL E 5 Solution Rationalizing Denominators Rationalize the denominator of each expression: 1 5 22 (a) (b) (c) 23 422 23 - 322 (a) The denominator contains the factor 23, so we multiply the numerator and denominator by 23 to obtain 1 23 = 1 # 23 23 23 = 23 1 232 2 = 23 3 (b) The denominator contains the factor 12, so we multiply the numerator and denominator by 12 to obtain 5 422 = 5 # 22 422 22 = 522 41 222 2 = 522 522 = 4#2 8 (c) The denominator contains the factor 23 - 322, so we multiply the numerator and denominator by 23 + 322 to obtain 22 23 - 322 = # 23 + 322 22 23 - 322 23 + 322 = 221 23 + 3222 1 232 - 13222 2 2 2 = Now Work 22 23 + 31 222 26 + 6 6 + 26 = = 3 - 18 - 15 15 PROBLEM 53 r 76 CHAPTER R Review 4 Simplify Expressions with Rational Exponents Radicals are used to define rational exponents. DEFINITION If a is a real number and n Ú 2 is an integer, then n a1>n = 2a (3) n provided that 2a exists. n Note that if n is even and a 6 0, then 2a and a1>n do not exist. EX AMPLE 6 Writing Expressions Containing Fractional Exponents as Radicals (a) 41>2 = 24 = 2 (b) 81>2 = 28 = 222 3 (c) 1 - 272 1>3 = 2 - 27 = - 3 DEFINITION r 3 3 (d) 161>3 = 2 16 = 22 2 If a is a real number and m and n are integers containing no common factors, with n Ú 2, then am>n = 2am = 1 2a2 m n n (4) n provided that 2a exists. We have two comments about equation (4): m must be in lowest terms, and n Ú 2 must be positive. n n n 2. In simplifying the rational expression am>n, either 2am or 1 2a2 m may be used, the choice depending on which is easier to simplify. Generally, taking the n root first, as in 1 2a2 m, is easier. 1. The exponent EX AMPLE 7 Using Equation (4) (a) 43>2 = 1 242 = 23 = 8 3 (b) 1 - 82 4>3 = 1 2 - 82 = 1 - 22 4 = 16 3 4 5 (c) 1322 -2>5 = 1 2 322 -2 = 2-2 = Now Work PROBLEM 1 3 (d) 256>4 = 253>2 = 1 2252 = 53 = 125 4 r 63 It can be shown that the Laws of Exponents hold for rational exponents. The next example illustrates using the Laws of Exponents to simplify. EX AMPLE 8 Simplifying Expressions Containing Rational Exponents Simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are positive. (a) 1x2>3 y2 1x -2 y2 1>2 (b) ¢ 2x1>3 y 2>3 ≤ -3 (c) ¢ 9x2 y1>3 x 1>3 y ≤ 1>2 SECTION R.8 nth Roots; Rational Exponents Solution (a) 1x2>3 y2 1x -2 y2 1>2 = 1x2>3 y2 3 1x -2 2 = x 2>3 = 1x 1>2 1>2 y 77 4 -1 1>2 yx y 2>3 # x-1 2 1y # y1>2 2 = x -1>3 y3>2 = (b) ¢ (c) ¢ 2x1>3 y2>3 ≤ -3 9x2 y1>3 x1>3 y ≤ = ¢ 1>2 y2>3 2x = ¢ Now Work y3>2 x1>3 3 ≤ = 1>3 9x2 - 11>32 y1 - 11>32 1y2>3 2 3 12x1>3 2 ≤ 1>2 PROBLEM 3 = ¢ = y2 23 1x1>3 2 9x5>3 y2>3 ≤ 1>2 3 = = y2 8x 91>2 1x5>3 2 1y2>3 2 1>2 1>2 = 3x5>6 y1>3 r 83 The next two examples illustrate some algebra that you will need to know for certain calculus problems. EXAMPL E 9 Writing an Expression as a Single Quotient Write the following expression as a single quotient in which only positive exponents appear. 1x2 + 12 Solution 1x2 + 12 1>2 + x# 1>2 + x# 1 2 1x + 12 -1>2 # 2x 2 1 2 x2 1>2 - 1>2 # 1x + 12 2 x = 1x2 + 12 + 1>2 2 1x2 + 12 = = = Now Work EX AM PL E 1 0 1>2 1x2 + 12 1x2 + 12 1>2 + x2 1>2 1x2 + 12 + x2 1x2 + 12 1>2 2x2 + 1 1x2 + 12 1>2 r 89 Factoring an Expression Containing Rational Exponents Factor and simplify: Solution PROBLEM 1x2 + 12 4 1>3 x 12x + 12 + 2x4>3 3 Begin by writing 2x4>3 as a fraction with 3 as the denominator. 4x1>3(2x + 1) 4x1>3(2x + 1) + 6x4>3 4 1>3 6x4>3 x 12x + 12 + 2x4>3 = + = 3 3 3 c 3 Add the two fractions 2x1>3[2(2x + 1) + 3x] 2x1>3(7x + 2) = 3 3 c c = 2 and x 1>3 are common factors Now Work PROBLEM 101 Simplify r 78 CHAPTER R Review Historical Note T he radical sign, 2 , was first used in print by Christoff Rudolff in 1525. It is thought to be the manuscript form of the letter r (for the Latin word radix = root), although this has not been quite conclusively confirmed. It took a long time for 2 to become the standard symbol for a square root and much longer to standardize 3 4 5 2 , 2, 2 , and so on. The indexes of the root were placed in every conceivable position, with 3 2 8, 3 8, and 2○ 3 4 all being variants for 2 8 . The notation 2 216 was popular for 2 16 . By the 1700s, the index had settled where we now put it. The bar on top of the present radical symbol, as follows, 2a2 + 2ab + b2 is the last survivor of the vinculum, a bar placed atop an expression to indicate what we would now indicate with parentheses. For example, 28 ab + c = a1b + c2 3 R.8 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages in red. 1. 1 - 32 2 = ; - 32 = 2. 216 = (pp. 21–24) ; 21 - 42 2 = (pp. 21–24) Concepts and Vocabulary n 3. In the symbol 2a, the integer n is called the 3 4. We call 2a the . of a. 5. Let n Ú 2 and m Ú 2 be integers, and let a and b be real numbers. Which of the following is not a property of radicals? Assume all radicals are defined. n (a) 2a a = n Ab 2b n n n n n 7. Which of the following phrases best defines like radicals? (a) Radical expressions that have the same index (b) Radical expressions that have the same radicand (c) Radical expressions that have the same index and the same radicand (d) Radical expressions that have the same variable 12 , 1 - 13 multiply both the numerator and the denominator by which n (b) 2a + b = 2a + 2b 8. To rationalize the denominator of the expression (d) 2am = 1 2a2 m n n (c) 2ab = 2a2b n 6. If a is a real number and n Ú 2 is an integer, then which of n the following expressions is equivalent to 2a, provided that it exists? 1 (a) a-n (b) an (c) (d) a1>n an of the following? (b) 22 (a) 23 (c) 1 + 23 (d) 1 - 23 5 9. True or False 2 - 32 = - 2 4 10. True or False 2 1 - 32 4 = - 3 Skill Building In Problems 11–48, simplify each expression. Assume that all variables are positive when they appear. 3 11. 227 4 12. 2 16 3 13. 2 -8 15. 28 3 16. 2 54 17. 2 - 8x4 4 12 8 19. 2 x y 5 10 5 20. 2 x y 21. 23. 236x 24. 29x5 4 25. 2162x9 y12 3 26. 2 - 40x14 y10 27. 23x2 212x 28. 25x 220x3 3 29. 1 25 2 92 3 30. 1 2 3 2102 31. 13262 12222 32. 15282 1 - 3232 33. 322 + 422 34. 625 - 425 35. - 218 + 228 36. 2212 - 3227 37. 1 23 + 32 1 23 - 12 38. 1 25 - 22 1 25 + 32 3 3 39. 52 2 - 22 54 3 3 40. 92 24 - 2 81 41. 1 2x - 12 2 42. 1 2x + 252 2 3 3 43. 216x4 - 22x 4 4 44. 2 32x + 2 2x5 45. 28x3 - 3250x 3 3 3 47. 216x4 y - 3x22xy + 52 - 2xy4 3 14. 2 -1 3 4 4 18. 2 48x5 x9 y 7 B xy 22. 3 2 3 3xy2 B 81x4 y2 4 46. 3x29y + 4225y 3 48. 8xy - 225x2 y2 + 28x3 y3 SECTION R.8 nth Roots; Rational Exponents 79 In Problems 49–62, rationalize the denominator of each expression. Assume that all variables are positive when they appear. 49. 53. 57. 61. 1 2 50. 22 23 22 54. 5 - 22 5 55. 27 + 2 - 23 52. 25 2 - 25 56. 2 + 325 -3 58. 22 - 1 51. 23 59. 25 + 4 2x + h - 2x 62. 2x + h + 2x In Problems 63–78, simplify each expression. - 23 28 23 - 1 223 + 3 5 60. 3 22 -2 3 2 9 2x + h + 2x - h 2x + h - 2x - h 63. 82>3 64. 43>2 65. 1 - 272 1>3 66. 163>4 67. 163>2 68. 253>2 69. 9-3>2 70. 16 -3>2 9 3>2 71. a b 8 72. a 8 -3>2 73. a b 9 74. a 75. 1 - 10002 -1>3 76. - 25-1>2 27 2>3 b 8 77. a - 64 -2>3 b 125 8 -2>3 b 27 78. - 81-3>4 In Problems 79–86, simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are positive. 79. x3>4 x1>3 x -1>2 83. 1x2 y2 1>3 81. 1x3 y6 2 80. x2>3 x1>2 x -1>4 1xy2 2 1xy2 1>4 1x2 y2 2 1>2 2>3 1x2y2 3>4 84. x2>3 y2>3 85. 82. 1x4y8 2 1>3 116x2 y -1>3 2 3>4 1xy2 2 1>4 86. 3>4 14x -1 y1>3 2 3>2 1xy2 3>2 Applications and Extensions In Problems 87–100, expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and/or radicals appear. 87. x 11 + x2 + 211 + x2 1>2 1>2 89. 2x1x2 + 12 1>2 + x2 # 91. 24x + 3 # 1 22x - 5 21 + x - x # 95. 97. 1x2 - 12 1>2 - 1x2 - 12 1 524x + 3 x 7 5 92. x 7 -1 x 7 -4 94. 96. 1>2 x 6 - 1 or x 7 1 98. 2 + x1>2 x 7 0 3 2 8x + 1 32 1x - 22 3 2 + 1 1x + 12 -2>3 3 x 7 0 100. x ≠ -1 3 2 x - 2 242 18x + 12 3 2 x ≠ 2, x ≠ - 1 8 2x 22x2 + 1 x + 1 2 19 - x2 2 1>2 + x2 19 - x2 2 -1>2 9 - x2 -3 6 x 6 3 1x2 + 42 1>2 - x2 1x2 + 42 -1>2 x2 + 4 2x11 - x2 2 1>3 + - 2x2x 11 + x2 2 2x1>2 2x2 + 1 - x # x2 22x 1 + x 90. 1x + 12 1>3 + x # 1 x + 4 1 + x2 99. + 2x - 5 # 1x + 42 1>2 - 2x1x + 42 -1>2 x2 88. 1 2 1x + 12 -1>2 # 2x 2 221 + x 1 + x 93. x 7 -1 2 3 x 11 - x2 2 -2>3 3 11 - x2 2 2>3 x ≠ - 1, x ≠ 1 80 CHAPTER R Review In Problems 101–110, expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive exponents occur. 3 4 4>3 1>3 3>2 101. 1x + 12 + x # 1x + 12 1>2 x Ú -1 102. 1x2 + 42 + x # 1x2 + 42 # 2x 2 3 103. 6x1>2 1x2 + x2 - 8x3>2 - 8x1>2 105. 31x2 + 42 4>3 + x # 41x2 + 42 104. 6x1>2 12x + 32 + x3>2 # 8 x Ú 0 1>3 106. 2x13x + 42 4>3 + x2 # 413x + 42 1>3 # 2x 107. 413x + 52 1>3 12x + 32 3>2 + 313x + 52 4>3 12x + 32 1>2 x Ú 109. 3x -1>2 + 3 1>2 x 2 x Ú 0 3 3 108. 616x + 12 1>3 14x - 32 3>2 + 616x + 12 4>3 14x - 32 1>2 x Ú 2 4 110. 8x1>3 - 4x -2>3 x 7 0 x ≠ 0 In Problems 111–118, use a calculator to approximate each radical. Round your answer to two decimal places. 111. 22 115. 2 + 23 3 - 25 3 113. 2 4 112. 27 116. 25 - 2 22 + 4 117. 3 114. 2 -5 3 32 5 - 22 118. 23 3 223 - 2 4 22 119. Calculating the Amount of Gasoline in a Tank A Shell station stores its gasoline in underground tanks that are right circular cylinders lying on their sides. See the illustration. The volume V of gasoline in the tank (in gallons) is given by the formula V = 40h2 96 - 0.608 Ah where h is the height of the gasoline (in inches) as measured on a depth stick. (a) If h = 12 inches, how many gallons of gasoline are in the tank? (b) If h = 1 inch, how many gallons of gasoline are in the tank? 120. Inclined Planes The final velocity v of an object in feet per second (ft>sec) after it slides down a frictionless inclined plane of height h feet is v = 264h + v20 where v0 is the initial velocity (in ft/sec) of the object. (a) What is the final velocity v of an object that slides down a frictionless inclined plane of height 4 feet? Assume that the initial velocity is 0. (b) What is the final velocity v of an object that slides down a frictionless inclined plane of height 16 feet? Assume that the initial velocity is 0. (c) What is the final velocity v of an object that slides down a frictionless inclined plane of height 2 feet with an initial velocity of 4 ft/sec? v0 h v Problems 121 and 122 require the following information. Period of a Pendulum The period T, in seconds, of a pendulum of length l, in feet, may be approximated using the formula T = 2p l A 32 In Problems 121 and 122, express your answer both as a square root and as a decimal. 121. Find the period T of a pendulum whose length is 64 feet. 122. Find the period T of a pendulum whose length is 16 feet. Explaining Concepts: Discussion and Writing 123. Give an example to show that 2a2 is not equal to a. Use it to explain why 2a2 = 0 a 0 . ‘Are You Prepared?’ Answers 1. 9; - 9 2. 4; 4 Equations and Inequalities Financing a Purchase 1 Whenever we make a major purchase, such as an automobile auto au t mobile or a house, money from a lending we often need to finance the purchase by borrowing money institution, such as a bank. Have you ever wondered how how the bank terest determines the monthly payment? How much total interest will be paid over the course of the loan? What roles do th the he rate of interest and the length of the loan play? —See the Internet-based Chapter Project I— — A Look Ahead Chapter 1, Equations and Inequalities, reviews many topics covered in Intermediate Algebra. If your instructor decides to exclude complex numbers from the course, don’t be alarmed. The book has been designed so that the topic of complex numbers can be included or excluded without any confusion later on. Outline 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Linear Equations Quadratic Equations Complex Numbers; Quadratic Equations in the Complex Number System Radical Equations; Equations Quadratic in Form; Factorable Equations Solving Inequalities Equations and Inequalities Involving Absolute Value Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications Chapter Review Chapter Test Chapter Projects 81 81 82 CHAPTER 1 Equations and Inequalities 1.1 Linear Equations PREPARING FOR THIS SECTION Before getting started, review the following: r Properties of Real Numbers (Section R.1, pp. 9–13) r Domain of a Variable (Section R.2, p. 21) Now Work the ‘Are You Prepared?’ problems on page 90. OBJECTIVES 1 Solve a Linear Equation (p. 84) 2 Solve Equations That Lead to Linear Equations (p. 86) 3 Solve Problems That Can Be Modeled by Linear Equations (p. 87) An equation in one variable is a statement in which two expressions, at least one containing the variable, are set equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be true or false, depending on the value of the variable. Unless otherwise restricted, the admissible values of the variable are those in the domain of the variable. These admissible values of the variable, if any, that result in a true statement are called solutions, or roots, of the equation. To solve an equation means to find all the solutions of the equation. For example, the following are all equations in one variable, x: x + 5 = 9 x2 + 5x = 2x - 2 x2 - 4 = 0 x + 1 2x2 + 9 = 5 The first of these statements, x + 5 = 9, is true when x = 4 and false for any other choice of x. That is, 4 is a solution of the equation x + 5 = 9. We also say that 4 satisfies the equation x + 5 = 9, because, when 4 is substituted for x, a true statement results. Sometimes an equation will have more than one solution. For example, the equation x2 - 4 = 0 x + 1 has x = - 2 and x = 2 as solutions. Usually, we will write the solutions of an equation in set notation. This set is called the solution set of the equation. For example, the solution set of the equation x2 - 9 = 0 is 5 - 3, 36 . Some equations have no real solution. For example, x2 + 9 = 5 has no real solution, because there is no real number whose square, when added to 9, equals 5. An equation that is satisfied for every value of the variable for which both sides are defined is called an identity. For example, the equation 3x + 5 = x + 3 + 2x + 2 is an identity, because this statement is true for any real number x. One method for solving an equation is to replace the original equation by a succession of equivalent equations, equations having the same solution set, until an equation with an obvious solution is obtained. For example, consider the following succession of equivalent equations: 2x + 3 = 13 2x = 10 x = 5 We conclude that the solution set of the original equation is 5 56 . How do we obtain equivalent equations? In general, there are five ways. SECTION 1.1 Linear Equations 83 Procedures That Result in Equivalent Equations 1. Interchange the two sides of the equation: Replace 3 = x by x = 3 2. Simplify the sides of the equation by combining like terms, eliminating parentheses, and so on: 1x + 22 + 6 = 2x + 1x + 12 x + 8 = 3x + 1 Replace by 3. Add or subtract the same expression on both sides of the equation: 3x - 5 = 4 13x - 52 + 5 = 4 + 5 Replace by 4. Multiply or divide both sides of the equation by the same nonzero expression: 3x 6 = x ≠ 1 x - 1 x - 1 3x # 6 # 1x - 12 = 1x - 12 x - 1 x - 1 Replace by WARNING Squaring both sides of an equation does not necessarily lead to an equivalent equation. For example, x = 3 has one solution, but x 2 = 9 has two solutions, x = - 3 and x = 3. 5. If one side of the equation is 0 and the other side can be factored, then we may use the Zero-Product Property* and set each factor equal to 0: x1x - 32 = 0 x = 0 or x - 3 = 0 Replace by ■ Whenever it is possible to solve an equation in your head, do so. For example, The solution of 2x = 8 is x = 4. The solution of 3x - 15 = 0 is x = 5. Now Work EXAMPL E 1 PROBLEM 11 Solving an Equation Solve the equation: 3x - 5 = 4 Solution Replace the original equation by a succession of equivalent equations. 3x - 5 = 4 13x - 52 + 5 = 4 + 5 3x = 9 3x 9 = 3 3 x = 3 Add 5 to both sides. Simplify. Divide both sides by 3. Simplify. The last equation, x = 3, has the single solution 3. All these equations are equivalent, so 3 is the only solution of the original equation, 3x - 5 = 4. *The Zero-Product Property says that if ab = 0, then a = 0 or b = 0 or both equal 0. 84 CHAPTER 1 Equations and Inequalities Check: Check the solution by substituting 3 for x in the original equation. 3x - 5 = 3132 - 5 ≟ 9 - 5 ≟ 4 = 4 4 4 4 The solution checks. The solution set is 5 36 . Now Work PROBLEM r 25 Steps for Solving Equations STEP 1: STEP 2: STEP 3: STEP 4: List any restrictions on the domain of the variable. Simplify the equation by replacing the original equation by a succession of equivalent equations using the procedures listed earlier. If the result of Step 2 is a product of factors equal to 0, use the Zero-Product Property and set each factor equal to 0 (procedure 5). Check your solution(s). 1 Solve a Linear Equation Linear equations are equations such as 3x + 12 = 0 DEFINITION - 2x + 5 = 0 1 x - 23 = 0 2 A linear equation in one variable is an equation equivalent in form to ax + b = 0 where a and b are real numbers and a ≠ 0. Sometimes a linear equation is called a first-degree equation, because the left side is a polynomial in x of degree 1. It is relatively easy to solve a linear equation. The idea is to isolate the variable: ax + b = 0 ax = - b -b x = a a ≠ 0 Subtract b from both sides. Divide both sides by a, a ≠ 0. The linear equation ax + b = 0, a ≠ 0, has the single solution given by the b formula x = - . a EX AMPLE 2 Solving a Linear Equation Solve the equation: 1 1 1x + 52 - 4 = 12x - 12 2 3 SECTION 1.1 Linear Equations Solution 85 To clear the equation of fractions, multiply both sides by 6, the least common 1 1 multiple (LCM) of the denominators of the fractions and . 2 3 1 1 1x + 52 - 4 = 12x - 12 2 3 1 1 6c 1x + 52 - 4 d = 6c 12x - 12 d Multiply both sides by 6, the LCM of 2 and 3. 2 3 31x + 52 - 6 # 4 = 212x - 12 3x + 15 - 24 3x - 9 3x - 9 + 9 3x 3x - 4x -x x = = = = = = = 4x 4x 4x 4x + 4x + 7 -7 2 2 2 + 9 7 7 - 4x Use the Distributive Property on the left and the Associative Property on the right. Use the Distributive Property. Combine like terms. Add 9 to each side. Simplify. Subtract 4x from each side. Simplify. Multiply both sides by - 1. Check: Substitute - 7 for x in the expressions on the left and right sides of the original equation, and simplify. If the two expressions are equal, the solution checks. 1 1 1 1x + 52 - 4 = 1 - 7 + 52 - 4 = 1 - 22 - 4 = - 1 - 4 = - 5 2 2 2 1 1 1 1 12x - 12 = 3 21 - 72 - 14 = 1 - 14 - 12 = 1 - 152 = - 5 3 3 3 3 Since the two expressions are equal, the solution checks. The solution set is 5 - 76 . Now Work EXAMPL E 3 PROBLEM r 35 Solving a Linear Equation Using a Calculator Solve the equation: 2.78x + 2 = 54.06 17.931 Round the answer to two decimal places. Solution To avoid rounding errors, solve for x before using a calculator. 2.78x + 2 = 54.06 17.931 2 2 from each side. Subtract 17.931 17.931 2 54.06 17.931 x = Divide each side by 2.78. 2.78 2.78x = 54.06 - Now use your calculator. The solution, rounded to two decimal places, is 19.41. Check: Store the calculator solution 19.40592134 in memory, and proceed 2 to evaluate 2.78x + . 17.931 12.782 119.405921342 + Now Work PROBLEM 67 2 = 54.06 17.931 r 86 CHAPTER 1 Equations and Inequalities 2 Solve Equations That Lead to Linear Equations Solving an Equation That Leads to a Linear Equation EX AMPL E 4 Solve the equation: 12y + 12 1y - 12 = 1y + 52 12y - 52 Solution 12y + 12 1y - 12 2y2 - y - 1 -y - 1 -y - 6y y = 1y + 52 12y - 52 = = = = = 2y2 + 5y - 25 5y - 25 5y - 24 - 24 4 Multiply and combine like terms. Subtract 2y2 from each side. Add 1 to each side. Subtract 5y from each side. Divide both sides by - 6. Check: 12y + 12 1y - 12 = 3 2142 + 14 14 - 12 = 18 + 12 132 = 192 132 = 27 1y + 52 12y - 52 = 14 + 52 3 2142 - 54 = 192 18 - 52 = 192 132 = 27 The two expressions are equal, so the solution checks. The solution set is 5 46 . r Solving an Equation That Leads to a Linear Equation EX AMPLE 5 Solve the equation: Solution 3 1 7 = + x - 2 x - 1 1x - 12 1x - 22 First, notice that the domain of the variable is 5 x x ≠ 1, x ≠ 26 . Clear the equation of fractions by multiplying both sides by the least common multiple of the denominators of the three fractions, 1x - 12 1x - 22. 3 1 7 = + x - 2 x - 1 1x - 12 1x - 22 1x - 12 1x - 22 3 1 7 = 1x - 12 1x - 22 c + d x - 2 x - 1 1x - 12 1x - 22 3x - 3 = 1x - 12 1x - 22 1 7 + 1x - 12 1x - 22 x - 1 1x - 12 1x - 22 3x - 3 = 1x - 22 + 7 3x - 3 = x + 5 Multiply both sides by (x - 1)(x - 2); cancel on the left. Use the Distributive Property on each side; cancel on the right. Simplify. Combine like terms. Add 3 to each side; subtract x from each side. 2x = 8 x = 4 Divide by 2. Check: 3 3 3 = = x - 2 4 - 2 2 1 7 1 7 1 7 2 7 9 3 + = + = + # = + = = x - 1 1x - 12 1x - 22 4 - 1 14 - 12 14 - 22 3 3 2 6 6 6 2 The solution checks. The solution set is 5 46 . Now Work EX AMPLE 6 PROBLEM 61 An Equation with No Solution Solve the equation: 3x 3 + 2 = x - 1 x - 1 r SECTION 1.1 Linear Equations Solution First, note that the domain of the variable is 5 x x ≠ 16 . Since the two quotients in the equation have the same denominator, x - 1, simplify by multiplying both sides by x - 1. The resulting equation is equivalent to the original equation, since we are multiplying by x - 1, which is not 0. (Remember, x ≠ 1.) 3x 3 + 2 = x - 1 x - 1 3x 3 a + 2b # 1x - 12 = x - 1 x - 1 3x x - 1 NOTE Example 6 illustrates the appearance of an extraneous solution. This will be discussed more in Section 1.4. ■ # # 1x - 12 1x - 12 + 2 # 1x - 12 = 3 3x + 2x - 2 5x - 2 5x x = = = = Multiply both sides by x - 1; cancel on the right. Use the Distributive Property on the left side; cancel on the left. 3 3 5 1 Simplify. Combine like terms. Add 2 to each side. Divide both sides by 5. The solution appears to be 1. But recall that x = 1 is not in the domain of the variable, so this value must be discarded. The equation has no solution. The solution set is ∅. r Now Work EXAMPL E 7 87 PROBLEM 51 Converting to Fahrenheit from Celsius In the United States we measure temperature in both degrees Fahrenheit (°F) and 5 degrees Celsius (°C), which are related by the formula C = 1F - 322. What are the 9 Fahrenheit temperatures corresponding to Celsius temperatures of 0°, 10°, 20°, and 30°C? Solution We could solve four equations for F by replacing C each time by 0, 10, 20, and 30. 5 Instead, it is much easier and faster first to solve the equation C = 1F - 322 9 for F and then to substitute in the values of C. 5 1F - 322 9 9C = 51F - 322 C = 9C = 5F - 160 5F - 160 = 9C 5F = 9C + 160 9 F = C + 32 5 Multiply both sides by 9. Use the Distributive Property. Interchange sides. Add 160 to each side. Divide both sides by 5. Now do the required arithmetic. 0°C: F = 9 102 + 32 = 32°F 5 9 1102 + 32 = 50°F 5 9 20°C: F = 1202 + 32 = 68°F 5 9 30°C: F = 1302 + 32 = 86°F 5 10°C: F = NOTE The icon icon is a Model It! icon. It indicates that the discussion or problem involves modeling. ■ r 3 Solve Problems That Can Be Modeled by Linear Equations Although each situation has its unique features, we can provide an outline of the steps to follow when solving applied problems. 88 CHAPTER 1 Equations and Inequalities Steps for Solving Applied Problems STEP 1: NOTE It is a good practice to choose a variable that reminds you of the unknown. For example, use t for time. ■ STEP 2: STEP 3: STEP 4: STEP 5: EX AMPLE 8 Read the problem carefully, perhaps two or three times. Pay particular attention to the question being asked in order to identify what you are looking for. Identify any relevant formulas you may need (d = rt, A = pr 2, etc.). If you can, determine realistic possibilities for the answer. Assign a letter (variable) to represent what you are looking for, and, if necessary, express any remaining unknown quantities in terms of this variable. Make a list of all the known facts, and translate them into mathematical expressions. These may take the form of an equation (or, later, an inequality) involving the variable. The equation (or inequality) is called the model. If possible, draw an appropriately labeled diagram to assist you. Sometimes a table or chart helps. Solve the equation for the variable, and then answer the question, usually using a complete sentence. Check the answer with the facts in the problem. If it agrees, congratulations! If it does not agree, try again. Investments A total of $18,000 is invested, some in stocks and some in bonds. If the amount invested in bonds is half that invested in stocks, how much is invested in each category? Step-by-Step Solution Step 1: Determine what you are looking for. We are being asked to find the amount of two investments. These amounts must total $18,000. (Do you see why?) Step 2: Assign a variable to represent what you are looking for. If necessary, express any remaining unknown quantities in terms of this variable. If x equals the amount invested in stocks, then the rest of the money, 18,000 - x, is the amount invested in bonds. Step 3: Translate the English into mathematical statements. It may be helpful to draw a figure that represents the situation. Sometimes a table can be used to organize the information. Use the information to build your model. Set up a table: Step 4: Solve the equation and answer the original question. Amount in Stocks Amount in Bonds Reason x 18,000 - x Total invested is $18,000. We also know that: Total amount invested in bonds is 18,000 - x = 18,000 - x = one@half that in stocks 1 1x2 2 1 x 2 18,000 = x + 1 x 2 Add x to both sides. 3 x Simplify. 2 2 2 3 2 a b 18,000 = a b a xb Multiply both sides by . 3 3 3 2 12,000 = x Simplify. So $12,000 is invested in stocks, and $18,000 - $12,000 = $6000 is invested in bonds. 18,000 = Step 5: Check your answer with the facts presented in the problem. The total invested is $12,000 + $6000 = $18,000, and the amount in bonds, $6000, is half that in stocks, $12,000. Now Work PROBLEM 83 r SECTION 1.1 Linear Equations 89 Determining an Hourly Wage EXAMPL E 9 Shannon grossed $725 one week by working 52 hours. Her employer pays time-and-a-half for all hours worked in excess of 40 hours. With this information, can you determine Shannon’s regular hourly wage? Solution STEP 1: We are looking for an hourly wage. Our answer will be expressed in dollars per hour. STEP 2: Let x represent the regular hourly wage, measured in dollars per hour. Then 1.5x is the overtime hourly wage. STEP 3: Set up a table: Hours Worked Hourly Wage Salary Regular 40 x 40x Overtime 12 1.5x 12(1.5x) = 18x The sum of regular salary plus overtime salary will equal $725. From the table, 40x + 18x = 725. STEP 4: 40x + 18x = 725 58x = 725 x = 12.50 Shannon’s regular hourly wage is $12.50 per hour. STEP 5: Forty hours yields a salary of 40 112.502 = $500, and 12 hours of overtime yields a salary of 1211.52 112.502 = $225, for a total of $725. Now Work PROBLEM 85 r SUMMARY Steps for Solving a Linear Equation To solve a linear equation, follow these steps: STEP 1: List any restrictions on the variable. STEP 2: If necessary, clear the equation of fractions by multiplying both sides by the least common multiple (LCM) of the denominators of all the fractions. STEP 3: Remove all parentheses and simplify. STEP 4: Collect all terms containing the variable on one side and all remaining terms on the other side. STEP 5: Simplify and solve. STEP 6: Check your solution(s). Historical Feature S olving equations is among the oldest of mathematical activities, and efforts to systematize this activity determined much of the shape of modern mathematics. Consider the following problem and its solution using only words: Solve the problem of how many apples Jim has, given that “Bob’s five apples and Jim’s apples together make twelve apples” by thinking, “Jim’s apples are all twelve apples less Bob’s five apples” and then concluding, “Jim has seven apples.” The mental steps translated into algebra are 5 + x = 12 x = 12 - 5 = 7 The solution of this problem using only words is the earliest form of algebra. Such problems were solved exactly this way in Babylonia in 1800 BC. We know almost nothing of mathematical work before this date, although most authorities believe the sophistication of the earliest known texts indicates that there must have been a long period of previous development. The method of writing out equations in words persisted for thousands of years, and although it now seems extremely cumbersome, it was used very effectively by many generations of mathematicians. The Arabs developed a good deal of the theory of cubic equations while writing out all the equations in words. About AD 1500, the tendency to abbreviate words in the written equations began to lead in the direction of modern notation; for example, the Latin word et (meaning “and”) developed into the plus sign, + . Although the occasional use of letters to represent variables dates back to AD 1200, the practice did not become common until about AD 1600. Development thereafter was rapid, and by 1635 algebraic notation did not differ essentially from what we use now. 90 CHAPTER 1 Equations and Inequalities 1.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. x 3. The domain of the variable in the expression is x - 4 . (p. 21) 1. The fact that 21x + 32 = 2x + 6 is attributable to the Property. (pp. 9–13) 2. The fact that 3x = 0 implies that x = 0 is a result of the Property. (pp. 9–13) Concepts and Vocabulary 8. True or False Some equations have no solution. 4. True or False Multiplying both sides of an equation by any number results in an equivalent equation. 9. An admissible value for the variable that makes the equation of the equation. a true statement is called a(n) (a) identity (b) solution (c) degree (d) model 5. An equation that is satisfied for every value of the variable for which both sides are defined is called a(n) . 6. An equation of the form ax + b = 0 is called a(n) equation or a(n) equation. 10. A chemist mixes 10 liters of a 20% solution with x liters of a 35% solution. Which of the following expressions represents the total number of liters in the mixture? 35 (a) x (b) 20 - x (c) (d) 10 + x x 7. True or False The solution of the equation 3x - 8 = 0 3 is . 8 Skill Building In Problems 11–18, mentally solve each equation. 11. 7x = 21 12. 6x = - 24 13. 3x + 15 = 0 15. 2x - 3 = 0 16. 3x + 4 = 0 17. 14. 6x + 18 = 0 1 5 x = 3 12 18. 2 9 x = 3 2 In Problems 19–66, solve each equation. 19. 3x + 4 = x 20. 2x + 9 = 5x 21. 2t - 6 = 3 - t 22. 5y + 6 = - 18 - y 23. 6 - x = 2x + 9 24. 3 - 2x = 2 - x 25. 3 + 2n = 4n + 7 26. 6 - 2m = 3m + 1 27. 213 + 2x2 = 31x - 42 28. 312 - x2 = 2x - 1 29. 8x - 13x + 22 = 3x - 10 30. 7 - 12x - 12 = 10 31. 1 1 3 x + 2 = - x 2 2 2 34. 1 - 1 x = 6 2 37. 0.9t = 0.4 + 0.1t 32. 2 1 x = 2 - x 3 3 33. 1 3 x - 5 = x 2 4 35. 2 1 1 p = p + 3 2 3 36. 1 1 4 - p = 2 3 3 39. x + 2 x + 1 + = 2 3 7 4 5 - 5 = y 2y 38. 0.9t = 1 + t 40. 2x + 1 + 16 = 3x 3 41. 4 2 + = 3 y y 42. 43. 1 2 3 + = 2 x 4 44. 3 1 1 - = x 3 6 45. 1x + 72 1x - 12 = 1x + 12 2 46. 1x + 22 1x - 32 = 1x + 32 2 47. x12x - 32 = 12x + 12 1x - 42 48. x11 + 2x2 = 12x - 12 1x - 22 49. z1z2 + 12 = 3 + z3 50. w 14 - w 2 2 = 8 - w 3 51. 2 x + 3 = x - 2 x - 2 52. 2x -6 = - 2 x + 3 x + 3 53. 4 3 2x = 2 x + 2 x - 4 x - 4 54. 4 x 3 + = 2 x + 3 x - 9 x - 9 55. 3 x = x + 2 2 56. 3x = 2 x - 1 57. 5 3 = 2x - 3 x + 5 58. -3 -4 = x + 4 x + 6 59. 6t + 7 3t + 8 = 4t - 1 2t - 4 60. 8w + 5 4w - 3 = 10w - 7 5w + 7 2 2 SECTION 1.1 Linear Equations 91 61. -3 7 4 = + x - 2 x + 5 1x + 52 1x - 22 62. -4 1 1 + = 2x + 3 x-1 12x + 32 1x - 12 63. 2 3 5 + = y + 3 y - 4 y + 6 64. 4 -3 5 + = 5z - 11 2z - 3 5 - z 65. x + 3 -3 x - 2 = 2 x2 - 1 x - x x + x 66. x + 1 x + 4 -3 - 2 = 2 x2 + 2x x + x x + 3x + 2 In Problems 67–70, use a calculator to solve each equation. Round the solution to two decimal places. 67. 3.2x + 21.3 = 19.23 65.871 69. 14.72 - 21.58x = 68. 6.2x - 18 x + 2.4 2.11 19.1 = 0.195 83.72 70. 18.63x - 14 21.2 = x - 20 2.6 2.32 Applications and Extensions In Problems 71–74, solve each equation. The letters a, b, and c are constants. 71. ax - b = c, a ≠ 0 73. 72. 1 - ax = b, a ≠ 0 x x + = c, a ≠ 0, b ≠ 0, a ≠ - b a b 75. Find the number a for which x = 4 is a solution of the equation x + 2a = 16 + ax - 6a 74. a b + = c, c ≠ 0 x x 76. Find the number b for which x = 2 is a solution of the equation x + 2b = x - 4 + 2bx Problems 77–82 list some formulas that occur in applications. Solve each formula for the indicated variable. 77. Electricity 1 1 1 = + R R1 R2 78. Finance A = P 11 + rt2 for R for r 2 mv for R R 80. Chemistry PV = nRT for T a 81. Mathematics S = for r 1 - r 82. Mechanics v = - gt + v0 for t 79. Mechanics F = 83. Finance A total of $20,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds is to exceed that in CDs by $3000, how much will be invested in each type of investment? 84. Finance A total of $10,000 is to be divided between Sean and George, with George to receive $3000 less than Sean. How much will each receive? 85. Computing Hourly Wages Sandra, who is paid time-and-a-half for hours worked in excess of 40 hours, had gross weekly wages of $546 for 48 hours worked. What is her regular hourly rate? 86. Computing Hourly Wages Leigh is paid time-and-a-half for hours worked in excess of 40 hours and double-time for hours worked on Sunday. If Leigh had gross weekly wages of $627 for working 50 hours, 4 of which were on Sunday, what is her regular hourly rate? 87. Computing Grades Going into the final exam, which will count as two tests, Brooke has test scores of 80, 83, 71, 61, and 95. What score does Brooke need on the final in order to have an average score of 80? 88. Computing Grades Going into the final exam, which will count as two-thirds of the final grade, Mike has test scores of 86, 80, 84, and 90. What minimum score does Mike need on the final in order to earn a B, which requires an average score of 80? What does he need to earn an A, which requires an average of 90? 89. Business: Discount Pricing A builder of tract homes reduced the price of a model by 15%. If the new price is $425,000, what was its original price? How much can be saved by purchasing the model? 90. Business: Discount Pricing A car dealer, at a year-end clearance, reduces the list price of last year’s models by 15%. If a certain four-door model has a discounted price of $8000, what was its list price? How much can be saved by purchasing last year’s model? 91. Personal Finance: Concession Markup A movie theater marks up the candy it sells by 275%. If a box of candy sells for $3.00 at the theater, how much did the theater pay for the box? 92 CHAPTER 1 Equations and Inequalities 92. Personal Finance: Cost of a Car The suggested list price of a new car is $18,000. The dealer’s cost is 85% of list. How much will you pay if the dealer is willing to accept $100 over cost for the car? 93. Business: Theater Attendance The manager of the Coral Theater wants to know whether the majority of its patrons are adults or children. One day in July, 5200 tickets were sold and the receipts totaled $29,961. The adult admission is $7.50, and the children’s admission is $4.50. How many adult patrons were there? 94. Business: Discount Pricing A wool suit, discounted by 30% for a clearance sale, has a price tag of $399. What was the suit’s original price? 95. Geometry The perimeter of a rectangle is 60 feet. Find its length and width if the length is 8 feet longer than the width. Tyshira tracks her net calories 98. Counting Calories (calories taken in minus calories burned) as part of her fitness program. For one particular day, her net intake was 1480 calories. Her lunch calories were half her breakfast calories, and her dinner calories were 200 more than her breakfast calories. She ate 120 less calories in snacks than for breakfast, and she burned 700 calories by exercising on her elliptical. How many calories did she take in from snacks? 99. Sharing the Cost of a Pizza Judy and Tom agree to share the cost of an $18 pizza based on how much each ate. If Tom 2 ate the amount that Judy ate, how much should each pay? 3 [Hint: Some pizza may be left.] Tom’s portion 96. Geometry The perimeter of a rectangle is 42 meters. Find its length and width if the length is twice the width. 97. Counting Calories Herschel uses an app on his smartphone to keep track of his daily calories from meals. One day his calories from breakfast were 125 more than his calories from lunch, and his calories from dinner were 300 less than twice his calories from lunch. If his total caloric intake from meals was 2025, determine his calories for each meal. Judy’s portion Explaining Concepts: Discussion and Writing 100. What Is Wrong? One step in the following list contains an error. Identify it and explain what is wrong. x = 2 3x - 2x = 2 3x = 2x + 2 x2 + 3x = x2 + 2x + 2 x + 3x - 10 = x2 + 2x - 8 2 1x - 22 1x + 52 = 1x - 22 1x + 42 x + 5 = x + 4 1 = 0 (1) (2) (3) (4) (5) (6) (7) (8) 101. The equation 8 + x 5 + 3 = x + 3 x + 3 has no solution, yet when we go through the process of solving it, we obtain x = - 3. Write a brief paragraph to explain what causes this to happen. 102. Make up an equation that has no solution and give it to a fellow student to solve. Ask the fellow student to write a critique of your equation. ‘Are You Prepared?’ Answers 1. Distributive 2. Zero-Product 3. 5x x ≠ 46 1.2 Quadratic Equations PREPARING FOR THIS SECTION Before getting started, review the following: r Factoring Polynomials (Section R.5, pp. 49–55) r Zero-Product Property (Section R.1, p. 13) r Square Roots (Section R.2, pp. 23–24) r Complete the Square (Section R.5, p. 56) Now Work the ‘Are You Prepared?’ problems on page 101. OBJECTIVES 1 2 3 4 Solve a Quadratic Equation by Factoring (p. 93) Solve a Quadratic Equation by Completing the Square (p. 95) Solve a Quadratic Equation Using the Quadratic Formula (p. 96) Solve Problems That Can Be Modeled by Quadratic Equations (p. 99) SECTION 1.2 Quadratic Equations 93 Quadratic equations are equations such as 2x2 + x + 8 = 0 3x2 - 5x + 6 = 0 x2 - 9 = 0 DEFINITION A quadratic equation is an equation equivalent to one of the form ax2 + bx + c = 0 (1) where a, b, and c are real numbers and a ≠ 0. A quadratic equation written in the form ax2 + bx + c = 0 is said to be in standard form. Sometimes, a quadratic equation is called a second-degree equation because, when it is in standard form, the left side is a polynomial of degree 2. We shall discuss three ways of solving quadratic equations: by factoring, by completing the square, and by using the quadratic formula. 1 Solve a Quadratic Equation by Factoring When a quadratic equation is written in standard form, it may be possible to factor the expression on the left side into the product of two first-degree polynomials. Then, by using the Zero-Product Property and setting each factor equal to 0, the resulting linear equations can be solved to obtain the solutions of the quadratic equation. EXAMPL E 1 Solving a Quadratic Equation by Factoring Solve the equations: (a) x2 + 6x = 0 Solution (b) 2x2 = x + 3 (a) The equation is in the standard form specified in equation (1). The left side may be factored as x2 + 6x = 0 x(x + 6) = 0 Factor. Using the Zero-Product Property, set each factor equal to 0 and then solve the resulting first-degree equations. x = 0 x = 0 or or The solution set is 5 0, - 66 . x + 6 = 0 x = -6 Zero-Product Property Solve. (b) Place the equation 2x2 = x + 3 in standard form by subtracting x and 3 from both sides. 2x2 = x + 3 2 2x - x - 3 = 0 Subtract x and 3 from both sides. The left side may now be factored as 12x - 32 1x + 12 = 0 Factor. so that 2x - 3 = 0 or 3 x = 2 3 The solution set is e - 1, f. 2 x + 1 = 0 x = -1 Zero-Product Property Solve. r 94 CHAPTER 1 Equations and Inequalities When the left side factors into two linear equations with the same solution, the quadratic equation is said to have a repeated solution. This solution is also called a root of multiplicity 2, or a double root. EX AMPLE 2 Solving a Quadratic Equation by Factoring Solve the equation: 9x2 - 6x + 1 = 0 Solution This equation is already in standard form, and the left side can be factored. 9x2 - 6x + 1 = 0 13x - 12 13x - 12 = 0 so x = 1 3 or x = 1 3 1 1 This equation has only the repeated solution . The solution set is e f. 3 3 Now Work PROBLEMS 13 AND r 23 The Square Root Method Suppose that we wish to solve the quadratic equation x2 = p (2) where p Ú 0 is a nonnegative number. Proceeding as in the earlier examples, x2 - p = 0 Put in standard form. 1x - 1p2 1x + 1p2 = 0 Factor (over the real numbers). x = 1p or x = - 1p Solve. We have the following result: If x2 = p and p Ú 0, then x = 1p or x = - 1p . (3) When statement (3) is used, it is called the Square Root Method. In statement (3), note that if p 7 0 the equation x2 = p has two solutions, x = 1p and x = - 1p . We usually abbreviate these solutions as x = { 1p , which is read as “x equals plus or minus the square root of p.” For example, the two solutions of the equation x2 = 4 are x = { 24 Use the Square Root Method. and, since 24 = 2, we have x = {2 The solution set is 5 - 2, 26 . EX AMPLE 3 Solving a Quadratic Equation Using the Square Root Method Solve each equation. (a) x2 = 5 Solution (b) 1x - 22 2 = 16 (a) Use the Square Root Method to get x2 = 5 x = { 25 x = 25 or x = - 25 The solution set is 5 - 15, 156 . Use the Square Root Method. SECTION 1.2 Quadratic Equations 95 (b) Use the Square Root Method to get 1x - 22 2 = 16 x - 2 = x - 2 = x - 2 = x = { 216 Use the Square Root Method. {4 116 = 4 4 or x - 2 = - 4 6 or x = -2 The solution set is 5 - 2, 66 . Now Work PROBLEM r 33 2 Solve a Quadratic Equation by Completing the Square EXAMPL E 4 Solving a Quadratic Equation by Completing the Square Solve by completing the square: x2 + 5x + 4 = 0 Solution Always begin this procedure by rearranging the equation so that the constant is on the right side. x2 + 5x + 4 = 0 x2 + 5x = - 4 NOTE If the coefficient of the square term is not 1, divide both sides by the coefficient of the square term before attempting to complete the square. ■ Since the coefficient of x2 is 1, complete the square on the left side by adding 2 1 25 a # 5b = . Of course, in an equation, whatever is added to the left side must 2 4 25 also be added to the right side. So add to both sides. 4 25 25 25 x2 + 5x + = -4 + Add to both sides. 4 4 4 5 2 9 ax + b = Factor the left side. 2 4 5 9 x + = { Use the Square Root Method. 74 2 3 9 5 19 3 x + = { = = 2 2 A4 2 14 5 3 x = - { 2 2 5 3 5 3 x = - + = - 1 or x = - - = - 4 2 2 2 2 The solution set is 5 - 4, - 16 . T H E S O LU T I O N O F T H E E Q UAT I O N I N E X A M P L E r 4 B E O BTA I N E D BY FAC TO R I N G . R E W O R K E X A M P L E FAC TO R I N G . CAN ALSO 4 USING The next example illustrates an equation that cannot be solved by factoring. EXAMPL E 5 Solving a Quadratic Equation by Completing the Square Solve by completing the square: 2x2 - 8x - 5 = 0 Solution First, rewrite the equation so that the constant is on the right side. 2x2 - 8x - 5 = 0 2x2 - 8x = 5 Add 5 to both sides. 96 CHAPTER 1 Equations and Inequalities Next, divide both sides by 2 so that the coefficient of x2 is 1. (This enables us to complete the square at the next step.) x2 - 4x = 5 2 Finally, complete the square by adding 4 to both sides. 5 x2 - 4x + 4 = + 4 Add 4 to both sides. 2 13 1x - 22 2 = Factor on the left; simplify on the right. 2 13 x - 2 = { Use the Square Root Method. A2 x - 2 = { 226 2 x = 2 { NOTE If we wanted an approximation, say rounded to two decimal places, of these solutions, we would use a calculator to get { - 0.55, 4.55}. ■ The solution set is e 2 - Now Work 13 113 113 = = A 2 12 12 # 12 12 = 126 2 226 2 226 226 ,2 + f. 2 2 PROBLEM r 37 3 Solve a Quadratic Equation Using the Quadratic Formula We can use the method of completing the square to obtain a general formula for solving any quadratic equation ax2 + bx + c = 0 NOTE There is no loss in generality to assume that a 7 0, since if a 6 0 we can multiply by - 1 to obtain an equivalent equation with a positive leading coefficient. ■ a ≠ 0 As in Examples 4 and 5, rearrange the terms as ax2 + bx = - c a 7 0 Since a 7 0, divide both sides by a to get x2 + c b x = a a Now the coefficient of x2 is 1. To complete the square on the left side, add the square 1 of of the coefficient of x; that is, add 2 1 b 2 b2 a # b = 2 a 4a2 to both sides. Then x2 + b b2 b2 c x + = 2 a a 4a 4a2 ax + b2 - 4ac b 2 b = 2a 4a2 b2 b2 c 4ac b 2 - 4ac - = = 2 2 2 a 4a 4a 4a 4a 2 Provided that b2 - 4ac Ú 0, we can now use the Square Root Method to get x + b2 - 4ac b = { 2a B 4a2 b { 2b2 - 4ac x + = 2a 2a The square root of a quotient equals the quotient of the square roots. Also, 24a2 = 2a since a 7 0. (4) SECTION 1.2 Quadratic Equations x = = b 2b2 - 4ac { 2a 2a - b { 2b2 - 4ac 2a Add - 97 b to both sides. 2a Combine the quotients on the right. What if b2 - 4ac is negative? Then equation (4) states that the left expression (a real number squared) equals the right expression (a negative number). Since this is impossible for real numbers, we conclude that if b2 - 4ac 6 0, the quadratic equation has no real solution. (We discuss quadratic equations for which the quantity b2 - 4ac 6 0 in detail in the next section.) THEOREM Quadratic Formula Consider the quadratic equation ax2 + bx + c = 0 a ≠ 0 If b2 - 4ac 6 0, this equation has no real solution. If b2 - 4ac Ú 0, the real solution(s) of this equation is (are) given by the quadratic formula: x = - b { 2b2 - 4ac 2a (5) The quantity b2 − 4ac is called the discriminant of the quadratic equation, because its value tells us whether the equation has real solutions. In fact, it also tells us how many solutions to expect. Discriminant of a Quadratic Equation For a quadratic equation ax2 + bx + c = 0, a ≠ 0: 1. If b2 - 4ac 7 0, there are two unequal real solutions. 2. If b2 - 4ac = 0, there is a repeated solution, a double root. 3. If b2 - 4ac 6 0, there is no real solution. When asked to find the real solutions, of a quadratic equation, always evaluate the discriminant first to see if there are any real solutions. EXAMPL E 6 Solving a Quadratic Equation Using the Quadratic Formula Use the quadratic formula to find the real solutions, if any, of the equation 3x2 - 5x + 1 = 0 Solution The equation is in standard form, so compare it to ax2 + bx + c = 0 to find a, b, and c. 3x2 - 5x + 1 = 0 ax2 + bx + c = 0 a = 3, b = - 5, c = 1 98 CHAPTER 1 Equations and Inequalities With a = 3, b = - 5, and c = 1, evaluate the discriminant b2 - 4ac. b2 - 4ac = 1 - 52 2 - 4132 112 = 25 - 12 = 13 Since b2 - 4ac 7 0, there are two real solutions, which can be found using the quadratic formula. x = - 1 - 52 { 213 - b { 2b2 - 4ac 5 { 213 = = 2a 2132 6 The solution set is e EX AMPLE 7 5 - 213 5 + 213 , f. 6 6 r Solving a Quadratic Equation Using the Quadratic Formula Use the quadratic formula to find the real solutions, if any, of the equation 25 2 x - 30x + 18 = 0 2 Solution The equation is given in standard form. However, to simplify the arithmetic, clear the fraction. 25 2 x - 30x + 18 = 0 2 25x2 - 60x + 36 = 0 Clear fraction; multiply by 2. ax2 + bx + c = 0 Compare to standard form. With a = 25, b = - 60, and c = 36, evaluate the discriminant. b2 - 4ac = 1 - 602 2 - 41252 1362 = 3600 - 3600 = 0 The equation has a repeated solution, which is found by using the quadratic formula. 60 { 20 60 6 - b { 2b2 - 4ac = = = 2a 50 50 5 x = 6 The solution set is e f. 5 EX AMPLE 8 r Solving a Quadratic Equation Using the Quadratic Formula Use the quadratic formula to find the real solutions, if any, of the equation 3x2 + 2 = 4x Solution The equation, as given, is not in standard form. 3x2 + 2 = 4x 3x2 - 4x + 2 = 0 Put in standard form. ax + bx + c = 0 Compare to standard form. 2 With a = 3, b = - 4, and c = 2, the discriminant is b2 - 4ac = 1 - 42 2 - 4132 122 = 16 - 24 = - 8 Since b2 - 4ac 6 0, the equation has no real solution. Now Work PROBLEMS 47 AND 57 r SECTION 1.2 Quadratic Equations EXAMPL E 9 Solving a Quadratic Equation Using the Quadratic Formula Find the real solutions, if any, of the equation: 9 + Solution 99 2 3 - 2 = 0, x ≠ 0 x x In its present form, the equation 9 + 3 2 - 2 = 0 x x is not a quadratic equation. However, it can be transformed into one by multiplying each side by x2. The result is 9x2 + 3x - 2 = 0 Although we multiplied each side by x2, we know that x2 ≠ 0 (do you see why?), so this quadratic equation is equivalent to the original equation. Using a = 9, b = 3, and c = - 2, the discriminant is b2 - 4ac = 32 - 4192 1 - 22 = 9 + 72 = 81 Since b2 - 4ac 7 0, the new equation has two real solutions. x = - b { 2b2 - 4ac - 3 { 281 -3 { 9 = = 2a 2192 18 x = -3 + 9 6 1 = = 18 18 3 or x = -3 - 9 - 12 2 = = 18 18 3 2 1 The solution set is e - , f. 3 3 r SUMMARY Procedure for Solving a Quadratic Equation To solve a quadratic equation, first put it in standard form: ax2 + bx + c = 0 Then: STEP 1: Identify a, b, and c. STEP 2: Evaluate the discriminant, b2 - 4ac. STEP 3: (a) If the discriminant is negative, the equation has no real solution. (b) If the discriminant is zero, the equation has one real solution, a double root. (c) If the discriminant is positive, the equation has two distinct real solutions. If you can easily spot factors, use the factoring method to solve the equation. Otherwise, use the quadratic formula or the method of completing the square. 4 Solve Problems That Can Be Modeled by Quadratic Equations Many applied problems require the solution of a quadratic equation. Let’s look at one that you will probably see again in a slightly different form if you study calculus. EX AM PL E 1 0 Constructing a Box From each corner of a square piece of sheet metal, remove a square of side 9 centimeters. Turn up the edges to form an open box. If the box is to hold 144 cubic centimeters (cm3), what should be the dimensions of the piece of sheet metal? 100 CHAPTER 1 Equations and Inequalities Solution Use Figure 1 as a guide. We have labeled by x the length of a side of the square piece of sheet metal. The box will be of height 9 centimeters, and its square base will measure x - 18 on each side. The volume V 1Length * Width * Height2 of the box is therefore V = 1x - 182 1x - 182 # 9 = 91x - 182 2 x cm 9 cm 9 cm 9 cm 9 cm x 18 9 cm 9 cm x 18 9 cm x cm x 18 9 cm x 18 Volume 9(x 18)(x 18) 9 cm Figure 1 Since the volume of the box is to be 144 cm3, we have Check: If we take a piece of sheet metal 22 centimeters by 22 centimeters, cut out a 9-centimeter square from each corner, and fold up the edges, we get a box whose dimensions are 9 by 4 by 4, with volume 9 * 4 * 4 = 144 cm3, as required. 91x - 182 2 1x - 182 2 x - 18 x x = = = = = 144 V = 144 16 Divide each side by 9. {4 Use the Square Root Method. 18 { 4 22 or x = 14 Discard the solution x = 14 (do you see why?) and conclude that the sheet metal should be 22 centimeters by 22 centimeters. r Now Work PROBLEM 99 Historical Feature P roblems using quadratic equations are found in the oldest known mathematical literature. Babylonians and Egyptians were solving such problems before 1800 BC. Euclid solved quadratic equations geometrically in his Data (300 BC), and the Hindus and Arabs gave rules for solving any quadratic equation with real roots. Because negative numbers were not freely used before AD 1500, there were several different types of quadratic equations, each with its own rule. Thomas Harriot (1560–1621) introduced the method of factoring to obtain solutions, and François Viète (1540–1603) introduced a method that is essentially completing the square. Until modern times it was usual to neglect the negative roots (if there were any), and equations involving square roots of negative quantities were regarded as unsolvable until the 1500s. Historical Problems 1. One of al-Khwǎrízmí solutions Solve x2 + 12x = 85 by drawing the square shown. The area of the four white rectangles and the yellow square is x2 + 12x. We then set this expression equal to 85 to get the equation x2 + 12x = 85. If we add the four blue x 3 x Area = x 2 3 Area = 3x u2 + (2z + 12)u + (z2 + 12z - 85) = 0 3 x x 3 3 3 2. Viète’s method Solve x2 + 12x - 85 = 0 by letting x = u + z. Then (u + z)2 + 12(u + z) - 85 = 0 3 3 squares, we will have a larger square of known area. Complete the solution. Now select z so that 2z + 12 = 0 and finish the solution. 3. Another method to get the quadratic formula Look at equation (4) 2b2 - 4ac 2 on page 96. Rewrite the right side as a b and then 2a subtract it from each side. The right side is now 0 and the left side is a difference of two squares. If you factor this difference of two squares, you will easily be able to get the quadratic formula, and moreover, the quadratic expression is factored, which is sometimes useful. SECTION 1.2 Quadratic Equations 101 1.2 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 4. True or False 2x2 = 0 x 0 . (pp. 23–24) 1. Factor: x2 - 5x - 6 (pp. 49–55) 2 2. Factor: 2x - x - 3 (pp. 49–55) 5. Complete the square of x2 + 5x. Factor the new expression. (p. 56) 3. The solution set of the equation 1x - 32 13x + 52 = 0 is . (p. 13) Concepts and Vocabulary 9. A quadratic equation is sometimes called a (a) first-degree (b) second-degree (c) third-degree (d) fourth-degree 6. The quantity b2 - 4ac is called the of a quadratic equation. If it is , the equation has no real solution. equation. 10. Which of the following quadratic equations is in standard form? (a) x2 - 7x = 5 (b) 9 = x2 (c) (x + 5)(x - 4) = 0 (d) 0 = 5x2 - 6x - 1 7. True or False Quadratic equations always have two real solutions. 8. True or False If the discriminant of a quadratic equation is positive, then the equation has two solutions that are negatives of one another. Skill Building In Problems 11–30, solve each equation by factoring. 11. x2 - 9x = 0 12. x2 + 4x = 0 13. x2 - 25 = 0 14. x2 - 9 = 0 15. z2 + z - 6 = 0 16. v2 + 7v + 6 = 0 17. 2x2 - 5x - 3 = 0 18. 3x2 + 5x + 2 = 0 19. 3t 2 - 48 = 0 20. 2y2 - 50 = 0 21. x1x - 82 + 12 = 0 22. x1x + 42 = 12 23. 4x2 + 9 = 12x 24. 25x2 + 16 = 40x 25. 61p2 - 12 = 5p 26. 212u2 - 4u2 + 3 = 0 27. 6x - 5 = 6 x 28. x + 12 = 7 x 29. In Problems 31–36, solve each equation by the Square Root Method. 41x - 22 x - 3 + 3 -3 = x x1x - 32 30. 5 3 = 4 + x + 4 x - 2 31. x2 = 25 32. x2 = 36 33. 1x - 12 2 = 4 34. 1x + 22 2 = 1 35. 12y + 32 2 = 9 36. 13z - 22 2 = 4 In Problems 37–42, solve each equation by completing the square. 37. x2 + 4x = 21 40. x2 + 2 1 x - = 0 3 3 38. x2 - 6x = 13 41. 3x2 + x - 1 = 0 2 39. x2 - 1 3 x = 0 2 16 42. 2x2 - 3x - 1 = 0 In Problems 43–66, find the real solutions, if any, of each equation. Use the quadratic formula. 43. x2 - 4x + 2 = 0 44. x2 + 4x + 2 = 0 45. x2 - 4x - 1 = 0 46. x2 + 6x + 1 = 0 47. 2x2 - 5x + 3 = 0 48. 2x2 + 5x + 3 = 0 49. 4y2 - y + 2 = 0 50. 4t 2 + t + 1 = 0 51. 4x2 = 1 - 2x 52. 2x2 = 1 - 2x 53. 4x2 = 9x 54. 5x = 4x2 55. 9t 2 - 6t + 1 = 0 56. 4u2 - 6u + 9 = 0 57. 3 2 1 1 x - x - = 0 4 4 2 102 58. CHAPTER 1 Equations and Inequalities 2 2 x - x - 3 = 0 3 61. 2x1x + 22 = 3 64. 4 + 1 1 - 2 = 0 x x 59. 5 2 1 x - x = 3 3 60. 62. 3x1x + 22 = 1 65. 1 3 2 x - x = 5 5 63. 4 - 1 3x + = 4 x - 2 x 66. 2 1 - 2 = 0 x x 2x 1 + = 4 x - 3 x In Problems 67–72, find the real solutions, if any, of each equation. Use the quadratic formula and a calculator. Express any solutions rounded to two decimal places. 67. x2 - 4.1x + 2.2 = 0 68. x2 + 3.9x + 1.8 = 0 69. x2 + 23 x - 3 = 0 70. x2 + 22 x - 2 = 0 71. px2 - x - p = 0 72. px2 + px - 2 = 0 In Problems 73–78, use the discriminant to determine whether each quadratic equation has two unequal real solutions, a repeated real solution (a double root), or no real solution, without solving the equation. 73. 2x2 - 6x + 7 = 0 74. x2 + 4x + 7 = 0 75. 9x2 - 30x + 25 = 0 76. 25x2 - 20x + 4 = 0 77. 3x2 + 5x - 8 = 0 78. 2x2 - 3x - 7 = 0 Mixed Practice In Problems 79–92, find the real solutions, if any, of each equation. Use any method. 79. x2 - 5 = 0 80. x2 - 6 = 0 81. 16x2 - 8x + 1 = 0 82. 9x2 - 12x + 4 = 0 83. 10x2 - 19x - 15 = 0 84. 6x2 + 7x - 20 = 0 85. 2 + z = 6z2 86. 2 = y + 6y2 87. x2 + 22 x = 90. x2 + x = 1 88. 1 2 x = 22 x + 1 2 89. x2 + x = 4 91. 2 7x + 1 x + = 2 x - 2 x + 1 x - x - 2 92. 1 2 3x 1 4 - 7x + = 2 x + 2 x - 1 x + x - 2 Applications and Extensions 93. Pythagorean Theorem How many right triangles have a hypotenuse that measures 2x + 3 meters and legs that measure 2x - 5 meters and x + 7 meters? What are their dimensions? 94. Pythagorean Theorem How many right triangles have a hypotenuse that measures 4x + 5 inches and legs that measure 3x + 13 inches and x inches? What are the dimensions of the triangle(s)? 95. Dimensions of a Window The area of the opening of a rectangular window is to be 143 square feet. If the length is to be 2 feet more than the width, what are the dimensions? 96. Dimensions of a Window The area of a rectangular window is to be 306 square centimeters. If the length exceeds the width by 1 centimeter, what are the dimensions? 97. Geometry Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters. 98. Watering a Field An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 1250 square feet (see the figure). What is the shortest radius setting that can be used if the field is to be completely enclosed within the circle? 99. Constructing a Box An open box is to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal? 100. Constructing a Box Rework Problem 99 if the piece of sheet metal is a rectangle whose length is twice its width. 101. Physics A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of SECTION 1.2 Quadratic Equations 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 96 + 80t - 16t 2. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down? 102. Physics An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (in meters) of the object from the ground after t seconds is s = - 4.9t 2 + 20t. (a) When will the object be 15 meters above the ground? (b) When will it strike the ground? (c) Will the object reach a height of 100 meters? 103. Reducing the Size of a Candy Bar A jumbo chocolate bar with a rectangular shape measures 12 centimeters in length, 7 centimeters in width, and 3 centimeters in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 10%. To accomplish this reduction, management decides that the new bar should have the same 3-centimeter thickness, but the length and width of each should be reduced an equal number of centimeters. What should be the dimensions of the new candy bar? 104. Reducing the Size of a Candy Bar Rework Problem 103 if the reduction is to be 20%. 105. Constructing a Border around a Pool A circular pool measures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of 3 inches, how wide will the border be? (1 cubic yard = 27 cubic feet) See the illustration. x 10 ft 103 109. Comparing Tablets The screen size of a tablet is determined by the length of the diagonal of the rectangular screen. The 9.7-inch iPad AirTM comes in a 4:3 format, which means that the ratio of the length to the width of the rectangular screen is 4:3. What is the area of the iPad’s screen? What is the area of a 10-inch Google NexusTM if its screen is in a 16:10 format? Which screen is larger? (Hint: If x is the length 3 of a 4:3 format screen, then x is the width.) 4 iPad mini 4:3 Google Nexus 16:10 110. Comparing Tablets Refer to Problem 109. Find the screen area of a 7.9-inch iPad mini with RetinaTM in a 4:3 format, and compare it with an 8-inch Dell Venue ProTM if its screen is in a 16:9 format. Which screen is larger? 111. Field Design A football field is sloped from the center toward the sides for drainage. The height h, in feet, of the field, x feet from the side, is given by h = - 0.00025x2 + 0.04x. Find the height of the field a distance of 35 feet from the side. Round to the nearest tenth of a foot. 112. College Value The difference, d, in median earnings, in $1000s, between high school graduates and college graduates can be approximated by d = - 0.002x2 + 0.319x + 7.512, where x is the number of years after 1965. Based on this model, estimate to the nearest year when the difference in median earnings was $15,000. (Source: Current Population Survey) 113. Student Working A study found that a student’s GPA, g, is related to the number of hours worked each week, h, by the equation g = - 0.0006h2 + 0.015h + 3.04. Estimate the number of hours worked each week for a student with a GPA of 2.97. Round to the nearest whole hour. 106. Constructing a Border around a Pool Rework Problem 105 if the depth of the border is 4 inches. 114. Fraternity Purchase A fraternity wants to buy a new LED Smart TV that costs $1470. If 7 members of the fraternity are not able to contribute, the share for the remaining members increases by $5. How many members are in the fraternity? 107. Constructing a Border around a Garden A landscaper, who just completed a rectangular flower garden measuring 6 feet by 10 feet, orders 1 cubic yard of premixed cement, all of which is to be used to create a border of uniform width around the garden. If the border is to have a depth of 3 inches, how wide will the border be? (1 cubic yard = 27 cubic feet) 115. The sum of the consecutive integers 1, 2, 3, c, n is given 1 by the formula n1n + 12. How many consecutive integers, 2 starting with 1, must be added to get a sum of 703? 1 116. Geometry If a polygon of n sides has n1n - 32 diagonals, 2 how many sides will a polygon with 65 diagonals have? Is there a polygon with 80 diagonals? 10 ft 6 ft 108. Dimensions of a Patio A contractor orders 8 cubic yards of premixed cement, all of which is to be used to pour a patio that will be 4 inches thick. If the length of the patio is specified to be twice the width, what will be the patio dimensions? (1 cubic yard = 27 cubic feet) 117. Show that the sum of the roots of a quadratic equation b is - . a 118. Show that the product of the roots of a quadratic equation c is . a 119. Find k such that the equation kx2 + x + k = 0 has a repeated real solution. 120. Find k such that the equation x2 - kx + 4 = 0 has a repeated real solution. 104 CHAPTER 1 Equations and Inequalities 121. Show that the real solutions of the equation ax2 + bx + c = 0 are the negatives of the real solutions of the equation ax2 - bx + c = 0. Assume that b2 - 4ac Ú 0. 122. Show that the real solutions of the equation ax2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx2 + bx + a = 0. Assume that b2 - 4ac Ú 0. Explaining Concepts: Discussion and Writing 123. Which of the following pairs of equations are equivalent? Explain. (a) x2 = 9; x = 3 (b) x = 29; x = 3 (c) 1x - 12 1x - 22 = 1x - 12 2; x - 2 = x - 1 124. Describe three ways that you might solve a quadratic equation. State your preferred method; explain why you chose it. 125. Explain the benefits of evaluating the discriminant of a quadratic equation before attempting to solve it. 126. Create three quadratic equations: one having two distinct real solutions, one having no real solution, and one having exactly one real solution. 127. The word quadratic seems to imply four (quad), yet a quadratic equation is an equation that involves a polynomial of degree 2. Investigate the origin of the term quadratic as it is used in the expression quadratic equation. Write a brief essay on your findings. ‘Are You Prepared?’ Answers 1. 1x - 62 1x + 12 2. 12x - 32 1x + 12 5 3. e - , 3 f 3 4. True 5. x2 + 5x + 5 2 25 = ax + b 4 2 1.3 Complex Numbers; Quadratic Equations in the Complex Number System * PREPARING FOR THIS SECTION Before getting started, review the following: r Classification of Numbers (Section R.1, pp. 4–5) r Rationalizing Denominators (Section R.8, p. 75) Now Work the ‘Are You Prepared?’ problems on page 111. OBJECTIVES 1 Add, Subtract, Multiply, and Divide Complex Numbers (p. 105) 2 Solve Quadratic Equations in the Complex Number System (p. 109) Complex Numbers One property of a real number is that its square is nonnegative. For example, there is no real number x for which x2 = - 1 To remedy this situation, we introduce a new number called the imaginary unit. DEFINITION The imaginary unit, which we denote by i, is the number whose square is - 1. That is, i2 = - 1 This should not surprise you. If our universe were to consist only of integers, there would be no number x for which 2x = 1. This was remedied by introducing 1 2 numbers such as and , the rational numbers. If our universe were to consist only 2 3 of rational numbers, there would be no x whose square equals 2. That is, there would be no number x for which x2 = 2. To remedy this, we introduced numbers such as 3 12 and 2 5, the irrational numbers. Recall that the real numbers consist of the rational numbers and the irrational numbers. Now, if our universe were to consist only of real numbers, then there would be no number x whose square is - 1. To remedy this, we introduce the number i, whose square is - 1. *This section may be omitted without any loss of continuity. SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System 105 In the progression outlined, each time we encountered a situation that was unsuitable, a new number system was introduced to remedy the situation. The number system that results from introducing the number i is called the complex number system. DEFINITION Complex numbers are numbers of the form a + bi, where a and b are real numbers. The real number a is called the real part of the number a + bi; the real number b is called the imaginary part of a + bi; and i is the imaginary unit, so i 2 = - 1. For example, the complex number - 5 + 6i has the real part - 5 and the imaginary part 6. When a complex number is written in the form a + bi, where a and b are real numbers, it is in standard form. However, if the imaginary part of a complex number is negative, such as in the complex number 3 + 1 - 22i, we agree to write it instead in the form 3 - 2i. Also, the complex number a + 0i is usually written merely as a. This serves to remind us that the real numbers are a subset of the complex numbers. Similarly, the complex number 0 + bi is usually written as bi. Sometimes the complex number bi is called a pure imaginary number. 1 Add, Subtract, Multiply, and Divide Complex Numbers Equality, addition, subtraction, and multiplication of complex numbers are defined so as to preserve the familiar rules of algebra for real numbers. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Equality of Complex Numbers a + bi = c + di if and only if a = c and b = d (1) Two complex numbers are added by forming the complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imaginary parts. Sum of Complex Numbers 1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i (2) To subtract two complex numbers, use this rule: Difference of Complex Numbers 1a + bi2 - 1c + di2 = 1a - c2 + 1b - d2i EXAMPL E 1 (3) Adding and Subtracting Complex Numbers (a) 13 + 5i2 + 1 - 2 + 3i2 = 3 3 + 1 - 22 4 + 15 + 32i = 1 + 8i (b) 16 + 4i2 - 13 + 6i2 = 16 - 32 + 14 - 62i = 3 + 1 - 22i = 3 - 2i Now Work PROBLEM 15 r 106 CHAPTER 1 Equations and Inequalities Products of complex numbers are calculated as illustrated in Example 2. EX AMPLE 2 Multiplying Complex Numbers 15 + 3i2 # 12 + 7i2 = 5 # 12 + 7i2 + 3i12 + 7i2 = 10 + 35i + 6i + 21i 2 c c Distributive Property Distributive Property = 10 + 41i + 211 - 12 c i2 = - 1 = - 11 + 41i r Based on the procedure of Example 2, the product of two complex numbers is defined as follows: Product of Complex Numbers 1a + bi2 # 1c + di2 = 1ac - bd2 + 1ad + bc2i (4) Do not bother to memorize formula (4). Instead, whenever it is necessary to multiply two complex numbers, follow the usual rules for multiplying two binomials, as in Example 2, remembering that i 2 = - 1. For example, 12i2 12i2 = 4i 2 = 41 - 12 = - 4 12 + i2 11 - i2 = 2 - 2i + i - i 2 = 3 - i Now Work PROBLEM 21 Algebraic properties for addition and multiplication, such as the commutative, associative, and distributive properties, hold for complex numbers. The property that every nonzero complex number has a multiplicative inverse, or reciprocal, requires a closer look. DEFINITION If z = a + bi is a complex number, then its conjugate, denoted by z, is defined as NOTE The conjugate of a complex number can be found by changing the sign of the imaginary part. ■ z = a + bi = a - bi For example, 2 + 3i = 2 - 3i and - 6 - 2i = - 6 + 2i. EX A MPLE 3 Multiplying a Complex Number by Its Conjugate Find the product of the complex number z = 3 + 4i and its conjugate z. Solution Since z = 3 - 4i, we have zz = 13 + 4i2 13 - 4i2 = 9 - 12i + 12i - 16i 2 = 9 + 16 = 25 r The result obtained in Example 3 has an important generalization. THEOREM The product of a complex number and its conjugate is a nonnegative real number. That is, if z = a + bi, then zz = a2 + b2 (5) SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System 107 Proof If z = a + bi, then zz = 1a + bi2 1a - bi2 = a2 - 1bi2 2 = a2 - b2 i 2 = a2 + b2 ■ To express the reciprocal of a nonzero complex number z in standard form, 1 multiply the numerator and denominator of by z. That is, if z = a + bi is a nonzero z complex number, then 1 1 z z a - bi 1 = = # = = 2 z z z a + bi zz a + b2 c Use (5). = EXAMPL E 4 Writing the Reciprocal of a Complex Number in Standard Form Write Solution a b - 2 i 2 a + b a + b2 2 1 in standard form a + bi; that is, find the reciprocal of 3 + 4i. 3 + 4i The idea is to multiply the numerator and denominator by the conjugate of 3 + 4i, that is, by the complex number 3 - 4i. The result is 1 1 # 3 - 4i 3 - 4i 3 4 = = = i 3 + 4i 3 + 4i 3 - 4i 9 + 16 25 25 r To express the quotient of two complex numbers in standard form, multiply the numerator and denominator of the quotient by the conjugate of the denominator. EXAMPL E 5 Writing the Quotient of Two Complex Numbers in Standard Form Write each of the following in standard form. (a) Solution (a) (b) 1 + 4i 5 - 12i 2 - 3i 4 - 3i 1 + 4i # 5 + 12i 5 + 12i + 20i + 48i 2 1 + 4i = = 5 - 12i 5 - 12i 5 + 12i 25 + 144 - 43 + 32i 43 32 = = + i 169 169 169 2 - 3i 2 - 3i # 4 + 3i 8 + 6i - 12i - 9i 2 17 - 6i 17 6 = = = = i 4 - 3i 4 - 3i 4 + 3i 16 + 9 25 25 25 Now Work EXAMPL E 6 (b) PROBLEM r 29 Writing Other Expressions in Standard Form If z = 2 - 3i and w = 5 + 2i, write each of the following expressions in standard form. (a) z w (b) z + w (c) z + z 108 CHAPTER 1 Equations and Inequalities Solution (a) 12 - 3i2 15 - 2i2 z z#w 10 - 4i - 15i + 6i 2 = = # = w w w 15 + 2i2 15 - 2i2 25 + 4 = 4 19 4 - 19i = i 29 29 29 (b) z + w = 12 - 3i2 + 15 + 2i2 = 7 - i = 7 + i (c) z + z = 12 - 3i2 + 12 + 3i2 = 4 r The conjugate of a complex number has certain general properties that will be useful later. For a real number a = a + 0i, the conjugate is a = a + 0i = a - 0i = a. THEOREM The conjugate of a real number is the real number itself. Other properties that are direct consequences of the definition of the conjugate are given next. In each statement, z and w represent complex numbers. THEOREM The conjugate of the conjugate of a complex number is the complex number itself. 1z2 = z (6) The conjugate of the sum of two complex numbers equals the sum of their conjugates. z + w = z + w (7) The conjugate of the product of two complex numbers equals the product of their conjugates. z#w = z#w (8) The proofs of equations (6), (7), and (8) are left as exercises. Powers of i The powers of i follow a pattern that is useful to know. i1 i2 i3 i4 = = = = i -1 i2 # i = - 1 # i = - i i 2 # i 2 = 1 - 12 1 - 12 = 1 i5 i6 i7 i8 = = = = i4 # i = 1 # i = i i4 # i2 = - 1 i4 # i3 = - i i4 # i4 = 1 And so on. The powers of i repeat with every fourth power. EX AMPLE 7 Evaluating Powers of i (a) i 27 = i 24 # i 3 = 1i 4 2 (b) i 101 = i 100 # i 1 = 6 # i3 = 16 # i3 = - i 25 1i 4 2 # i = 125 # i = i r SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System EXAMPL E 8 Solution 109 Writing the Power of a Complex Number in Standard Form Write 12 + i2 3 in standard form. Use the special product formula for 1x + a2 3. 1x + a2 3 = x3 + 3ax2 + 3a2 x + a3 NOTE Another way to find (2 + i)3 is to multiply out (2 + i)2 (2 + i). ■ Using this special product formula, 12 + i2 3 = 23 + 3 # i # 22 + 3 # i 2 # 2 + i 3 = 8 + 12i + 61 - 12 + 1 - i2 r = 2 + 11i Now Work PROBLEM 43 2 Solve Quadratic Equations in the Complex Number System Quadratic equations with a negative discriminant have no real number solution. However, if we extend our number system to allow complex numbers, quadratic equations will always have a solution. Since the solution to a quadratic equation involves the square root of the discriminant, we begin with a discussion of square roots of negative numbers. DEFINITION WARNING In writing 1- N = 1N i, be sure to place i outside the 1 symbol. ■ EXAMPL E 9 If N is a positive real number, we define the principal square root of −N, denoted by 1- N , as 2 - N = 2N i where i is the imaginary unit and i 2 = - 1. Evaluating the Square Root of a Negative Number (a) 1- 1 = 11 i = i (b) 1- 4 = 14 i = 2i (c) 1- 8 = 18 i = 212 i EX AM PL E 1 0 r Solving Equations Solve each equation in the complex number system. (a) x2 = 4 Solution (b) x2 = - 9 (a) x2 = 4 x = { 24 = {2 The equation has two solutions, - 2 and 2. The solution set is 5 - 2, 26 . (b) x2 = - 9 x = { 2 - 9 = { 29 i = {3i The equation has two solutions, - 3i and 3i. The solution set is 5 - 3i, 3i6 . Now Work PROBLEMS 51 AND 55 r 110 CHAPTER 1 Equations and Inequalities WARNING When working with square roots of negative numbers, do not set the square root of a product equal to the product of the square roots (which can be done with positive real numbers). To see why, look at this calculation: We know that 1100 = 10. However, it is also true that 100 = 1 - 252 1 - 42, so 10 = 2100 = 21 - 252 1 - 42 = 2- 25 2- 4 = 1 225 i 2 1 24 i 2 = 15i2 12i2 = 10i 2 = - 10 c Here is the error. ■ Because we have defined the square root of a negative number, we can now restate the quadratic formula without restriction. THEOREM Quadratic Formula In the complex number system, the solutions of the quadratic equation ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0, are given by the formula x = EX A MPL E 11 - b { 2b2 - 4ac 2a (9) Solving a Quadratic Equation in the Complex Number System Solve the equation x2 - 4x + 8 = 0 in the complex number system. Solution Here a = 1, b = - 4, c = 8, and b2 - 4ac = 1 - 42 2 - 4112 182 = - 16. Using equation (9), we find that x = - 1 - 42 { 2 - 16 2(2 { 2i) 4 { 216 i 4 { 4i = = = = 2 { 2i 2112 2 2 2 The equation has two solutions: 2 - 2i and 2 + 2i. The solution set is 5 2 - 2i, 2 + 2i6 . Check: 2 + 2i: 2 - 2i: Now Work 12 + 2i2 2 - 412 + 2i2 + 8 = 4 + 8i + 4i 2 - 8 - 8i + 8 = = 12 - 2i2 2 - 412 - 2i2 + 8 = = PROBLEM 4 4 4 4 + - 4i 2 4 = 0 8i + 4i 2 - 8 + 8i + 8 4 = 0 r 61 The discriminant b2 - 4ac of a quadratic equation still serves as a way to determine the character of the solutions. Character of the Solutions of a Quadratic Equation In the complex number system, consider a quadratic equation ax2 + bx + c = 0 with real coefficients. 1. If b2 - 4ac 7 0, the equation has two unequal real solutions. 2. If b2 - 4ac = 0, the equation has a repeated real solution, a double root. 3. If b2 - 4ac 6 0, the equation has two complex solutions that are not real. The solutions are conjugates of each other. SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System 111 The third conclusion in the display is a consequence of the fact that if b2 - 4ac = - N 6 0, then by the quadratic formula, the solutions are x = - b + 2b2 - 4ac - b + 2- N - b + 2N i -b 2N = = = + i 2a 2a 2a 2a 2a and - b - 2b2 - 4ac - b - 2- N - b - 2N i -b 2N = = = i 2a 2a 2a 2a 2a which are conjugates of each other. x = Determining the Character of the Solutions of a Quadratic Equation EX AM PL E 1 2 Without solving, determine the character of the solutions of each equation. (a) 3x2 + 4x + 5 = 0 Solution (b) 2x2 + 4x + 1 = 0 (c) 9x2 - 6x + 1 = 0 (a) Here a = 3, b = 4, and c = 5, so b2 - 4ac = 16 - 4132 152 = - 44. The solutions are two complex numbers that are not real and are conjugates of each other. (b) Here a = 2, b = 4, and c = 1, so b2 - 4ac = 16 - 8 = 8. The solutions are two unequal real numbers. (c) Here a = 9, b = - 6, and c = 1, so b2 - 4ac = 36 - 4192 112 = 0. The solution is a repeated real number—that is, a double root. r Now Work PROBLEM 75 1.3 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Name the integers and the rational numbers in the set 6 b - 3, 0, 12, , p r. (pp. 4–5) 5 2. True or False Rational numbers and irrational numbers are in the set of real numbers. (pp. 4–5) 3 3. Rationalize the denominator of . (p. 75) 2 + 13 Concepts and Vocabulary 4. In the complex number 5 + 2i, the number 5 is called the part; the number 2 is called the part; the number i is called the . 5. True or False The conjugate of 2 + 5i is - 2 - 5i. 6. True or False All real numbers are complex numbers. 7. True or False If 2 - 3i is a solution of a quadratic equation with real coefficients, then - 2 + 3i is also a solution. 8. Which of the following is the principal square root of - 4? (a) - 2i (b) 2i (c) - 2 (d) 2 9. Which operation involving complex numbers requires the use of a conjugate? (a) division (b) multiplication (c) subtraction (d) addition 10. Powers of i repeat every (a) second (b) third power. (c) fourth (d) fifth Skill Building In Problems 11–48, perform the indicated operation, and write each expression in the standard form a + bi. 11. 12 - 3i2 + 16 + 8i2 12. 14 + 5i2 + 1 - 8 + 2i2 13. 1 - 3 + 2i2 - 14 - 4i2 14. 13 - 4i2 - 1 - 3 - 4i2 15. 12 - 5i2 - 18 + 6i2 16. 1 - 8 + 4i2 - 12 - 2i2 17. 312 - 6i2 18. - 412 + 8i2 19. 2i12 - 3i2 20. 3i1 - 3 + 4i2 21. 13 - 4i2 12 + i2 22. 15 + 3i2 12 - i2 23. 1 - 6 + i2 1 - 6 - i2 24. 1 - 3 + i2 13 + i2 25. 10 3 - 4i 26. 13 5 - 12i 29. 6 - i 1 + i 30. 2 + 3i 1 - i 27. 2 + i i 31. a 1 23 2 + ib 2 2 28. 2 - i - 2i 32. a 23 1 2 - ib 2 2 33. 11 + i2 2 34. 11 - i2 2 112 CHAPTER 1 Equations and Inequalities 35. i 23 36. i 14 37. i -15 38. i -23 40. 4 + i 3 41. 6i 3 - 4i 5 42. 4i 3 - 2i 2 + 1 43. 11 + i2 3 45. i 7 11 + i 2 2 46. 2i 4 11 + i 2 2 47. i 6 + i 4 + i 2 + 1 48. i 7 + i 5 + i 3 + i 39. i 6 - 5 44. 13i2 4 + 1 In Problems 49–54, perform the indicated operations, and express your answer in the form a + bi. 49. 2- 4 50. 2- 9 51. 2- 25 52. 2- 64 53. 213 + 4i2 14i - 32 54. 214 + 3i2 13i - 42 In Problems 55–74, solve each equation in the complex number system. 55. x2 + 4 = 0 56. x2 - 4 = 0 57. x2 - 16 = 0 58. x2 + 25 = 0 59. x2 - 6x + 13 = 0 60. x2 + 4x + 8 = 0 61. x2 - 6x + 10 = 0 62. x2 - 2x + 5 = 0 63. 8x2 - 4x + 1 = 0 64. 10x2 + 6x + 1 = 0 65. 5x2 + 1 = 2x 66. 13x2 + 1 = 6x 67. x2 + x + 1 = 0 68. x2 - x + 1 = 0 69. x3 - 8 = 0 70. x3 + 27 = 0 71. x4 = 16 72. x4 = 1 73. x4 + 13x2 + 36 = 0 74. x4 + 3x2 - 4 = 0 In Problems 75–80, without solving, determine the character of the solutions of each equation in the complex number system. 75. 3x2 - 3x + 4 = 0 76. 2x2 - 4x + 1 = 0 77. 2x2 + 3x = 4 78. x2 + 6 = 2x 79. 9x2 - 12x + 4 = 0 80. 4x2 + 12x + 9 = 0 81. 2 + 3i is a solution of a quadratic equation with real coefficients. Find the other solution. 82. 4 - i is a solution of a quadratic equation with real coefficients. Find the other solution. In Problems 83–86, z = 3 - 4i and w = 8 + 3i. Write each expression in the standard form a + bi. 83. z + z 85. zz 84. w - w 86. z - w Applications and Extensions 87. Electrical Circuits The impedance Z, in ohms, of a circuit element is defined as the ratio of the phasor voltage V, in volts, across the element to the phasor current I, in amperes, V through the elements. That is, Z = . If the voltage across a I circuit element is 18 + i volts and the current through the element is 3 - 4 i amperes, determine the impedance. 88. Parallel Circuits In an ac circuit with two parallel pathways, the total impedance Z, in ohms, satisfies the formula 1 1 1 = + , where Z1 is the impedance of the first pathway Z Z1 Z2 and Z2 is the impedance of the second pathway. Determine the total impedance if the impedances of the two pathways are Z1 = 2 + i ohms and Z2 = 4 - 3i ohms. 89. Use z = a + bi to show that z + z = 2a and z - z = 2bi. 90. Use z = a + bi to show that z = z. 91. Use z = a + bi and w = c + di to show that z + w = z + w. 92. Use z = a + bi and w = c + di to show that z # w = z # w. Explaining Concepts: Discussion and Writing 93. Explain to a friend how you would add two complex numbers and how you would multiply two complex numbers. Explain any differences between the two explanations. 94. Write a brief paragraph that compares the method used to rationalize the denominator of a radical expression and the method used to write the quotient of two complex numbers in standard form. 96. Explain how the method of multiplying two complex numbers is related to multiplying two binomials. 97. What Went Wrong? A student multiplied 2- 9 and 2- 9 as follows: 2- 9 # 2- 9 = 2( - 9)( - 9) = 281 95. Use an Internet search engine to investigate the origins of complex numbers. Write a paragraph describing what you find, and present it to the class. = 9 The instructor marked the problem incorrect. Why? ‘Are You Prepared?’ Answers 6 1. Integers: 5 - 3, 06; rational numbers: b - 3, 0, r 5 2. True 3. 3 1 2 - 23 2 SECTION 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations 113 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations PREPARING FOR THIS SECTION Before getting started, review the following: r Square Roots (Section R.2, pp. 23–24) r Factoring Polynomials (Section R.5, pp. 49–55) Now Work the ‘Are You Prepared?’ problems on page 117. r nth Roots; Rational Exponents (Section R.8, pp. 73–77) OBJECTIVES 1 Solve Radical Equations (p. 113) 2 Solve Equations Quadratic in Form (p. 114) 3 Solve Equations by Factoring (p. 116) 1 Solve Radical Equations When the variable in an equation occurs in a square root, cube root, and so on—that is, when it occurs in a radical—the equation is called a radical equation. Sometimes a suitable operation will change a radical equation to one that is linear or quadratic. A commonly used procedure is to isolate the most complicated radical on one side of the equation and then eliminate it by raising each side to a power equal to the index of the radical. Care must be taken, however, because apparent solutions that are not, in fact, solutions of the original equation may result. These are called extraneous solutions. Therefore, we need to check all answers when working with radical equations, and we check them in the original equation. EXAMPL E 1 Solving a Radical Equation 3 Find the real solutions of the equation: 2 2x - 4 - 2 = 0 Solution The equation contains a radical whose index is 3. Isolate it on the left side. 3 2 2x - 4 - 2 = 0 3 2 2x - 4 = 2 Add 2 to both sides. Now raise each side to the third power (the index of the radical is 3) and solve. 3 12 2x - 42 2x - 4 2x x 3 = = = = 23 8 12 6 Raise each side to the power 3. Simplify. Add 4 to both sides. Divide both sides by 2. 3 3 3 Check: 2 2162 - 4 - 2 = 2 12 - 4 - 2 = 2 8 - 2 = 2 - 2 = 0 The solution set is 5 66 . Now Work EXAMPL E 2 r PROBLEM 9 Solving a Radical Equation Find the real solutions of the equation: 2x - 1 = x - 7 Solution Square both sides since the index of a square root is 2. 2x - 1 - 122 x - 1 2 x - 15x + 50 1 2x = = = = x - 7 1x - 72 2 x2 - 14x + 49 0 Square both sides. Remove parentheses. Put in standard form. 114 CHAPTER 1 Equations and Inequalities 1x - 102 1x - 52 = 0 x = 10 or x = 5 Factor. Apply the Zero-Product Property and solve. Check: x = 10: 2x - 1 = 210 - 1 = 29 = 3 and x - 7 = 10 - 7 = 3 x = 5: 2x - 1 = 25 - 1 = 24 = 2 and x - 7 = 5 - 7 = - 2 The solution x = 5 does not check, so it is extraneous; the only solution of the equation is x = 10. The solution set is {10}. r Now Work PROBLEM 19 Sometimes it is necessary to raise each side to a power more than once in order to solve a radical equation. Solving a Radical Equation EX AMPLE 3 Find the real solutions of the equation: 22x + 3 - 2x + 2 = 2 First, isolate the more complicated radical expression (in this case, 12x + 3) on the left side. Solution 22x + 3 = 2x + 2 + 2 Now square both sides (the index of the radical on the left is 2). 1 22x + 32 = 2 2x + 3 = 1 2x 1 2x + 2 + 222 + 2 2 + 42x + 2 + 4 2 Square both sides. Multiply out. 2x + 3 = x + 2 + 42x + 2 + 4 Simplify. 2x + 3 = x + 6 + 42x + 2 Combine like terms. Because the equation still contains a radical, isolate the remaining radical on the right side and again square both sides. x - 3 = 42x + 2 1x - 32 2 = 16(x + 2) 2 x - 6x + 9 x - 22x - 23 1x - 232 1x + 12 x = 23 or x 2 = = = = 16x + 32 0 0 -1 Isolate the radical on the right side. Square both sides. Multiply out. Put in standard form. Factor. Apply the Zero-Product Property and solve. The original equation appears to have the solution set 5 - 1, 236 . However, we have not yet checked. Check: x = 23: 22x + 3 - 2x + 2 = 221232 + 3 - 223 + 2 = 249 - 225 = 7 - 5 = 2 x = - 1: 22x + 3 - 2x + 2 = 221 - 12 + 3 - 2 - 1 + 2 = 21 - 21 = 1 - 1 = 0 The equation has only one solution, 23; the solution - 1 is extraneous. The solution set is {23}. r Now Work PROBLEM 31 2 Solve Equations Quadratic in Form The equation x4 + x2 - 12 = 0 is not quadratic in x, but it is quadratic in x2. That is, if we let u = x2, we get u2 + u - 12 = 0, a quadratic equation. This equation can be solved for u, and in turn, by using u = x2, we can find the solutions x of the original equation. SECTION 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations 115 In general, if an appropriate substitution u transforms an equation into one of the form au2 + bu + c = 0 a ≠ 0 then the original equation is called an equation of the quadratic type or an equation quadratic in form. The difficulty of solving such an equation lies in the determination that the equation is, in fact, quadratic in form. After you are told an equation is quadratic in form, it is easy enough to see it, but some practice is needed to enable you to recognize such equations on your own. EXAMPL E 4 Solution Solving an Equation Quadratic in Form Find the real solutions of the equation: 1x + 22 2 + 111x + 22 - 12 = 0 For this equation, let u = x + 2. Then u2 = 1x + 22 2, and the original equation, 1x + 22 2 + 111x + 22 - 12 = 0 becomes u2 + 11u - 12 = 0 1u + 122 1u - 12 = 0 u = - 12 or u = 1 Let u = x + 2. Then u 2 = (x + 2)2. Factor. Solve. But we want to solve for x. Because u = x + 2, we have x + 2 = - 12 or x + 2 = 1 x = - 14 x = -1 Check: x = - 14: x = - 1: 1 - 14 + 22 2 + 111 - 14 + 22 - 12 = 1 - 122 2 + 111 - 122 - 12 = 144 - 132 - 12 = 0 1 - 1 + 22 2 + 111 - 1 + 22 - 12 = 1 + 11 - 12 = 0 The original equation has the solution set 5 - 14, - 16 . EXAMPL E 5 r Solving an Equation Quadratic in Form Find the real solutions of the equation: 1x2 - 12 + 1x2 - 12 - 12 = 0 2 Solution For the equation 1x2 - 12 + 1x2 - 12 - 12 = 0, 2 that u2 = 1x2 - 12 . Then the original equation, 2 let u = x2 - 1 so 1x2 - 12 + 1x2 - 12 - 12 = 0 2 becomes u2 + u - 12 = 0 1u + 42 1u - 32 = 0 u = - 4 or u = 3 2 Let u = x 2 - 1. Then u 2 = (x 2 - 12 . Factor. Solve. But remember that we want to solve for x. Because u = x2 - 1, we have x2 - 1 = - 4 or x2 - 1 = 3 x2 = - 3 x2 = 4 The first of these has no real solution; the second has the solution set 5 - 2, 26 . Check: x = - 2: x = 2: 1( - 2)2 - 12 2 + 1( - 2)2 - 12 - 12 = 9 + 3 - 12 = 0 1(2)2 - 12 2 + 1(2)2 - 12 - 12 = 9 + 3 - 12 = 0 The original equation has the solution set { - 2, 2}. r 116 CHAPTER 1 Equations and Inequalities EX AMPLE 6 Solving an Equation Quadratic in Form Find the real solutions of the equation: x + 21x - 3 = 0 Solution For the equation x + 21x - 3 = 0, let u = 1x. Then u2 = x, and the original equation, x + 21x - 3 = 0 becomes u2 + 2u - 3 = 0 Let u = 1x. Then u 2 = x. 1u + 32 1u - 12 = 0 Factor. u = - 3 or u = 1 Solve. Since u = 1x, we have 1x = - 3 or 1x = 1. The first of these, 1x = - 3, has no real solution, since the principal square root of a real number is never negative. The second, 1x = 1, has the solution x = 1. Check: 1 + 221 - 3 = 1 + 2 - 3 = 0 r The original equation has the solution set {1}. A N OT H E R M E T H O D F O R S O LV I N G E XA M P L E 6 WOULD B E TO T R E AT I T A S A R A D I C A L E Q UAT I O N . S O LV E I T T H I S WAY F O R P R A C T I C E . The idea should now be clear. If an equation contains an expression and that same expression squared, make a substitution for the expression. You may get a quadratic equation. Now Work PROBLEM 53 3 Solve Equations by Factoring We have already solved certain quadratic equations using factoring. Let’s look at examples of other kinds of equations that can be solved by factoring. EX AMPLE 7 Solving an Equation by Factoring Solve the equation: x4 = 4x2 Solution Begin by collecting all terms on one side. This results in 0 on one side and an expression to be factored on the other. x4 = 4x2 x4 - 4x2 = 0 x2 1x2 - 42 = 0 Factor. x = 0 or x - 4 = 0 Apply the Zero-Product Property. x = 0 x = 4 2 x = 0 or x = 2: 2 or x = - 2 or x = 2 Check: x = - 2: x = 0: 2 1 - 22 4 = 16 and 41 - 22 2 = 16 4 0 4 2 The solution set is 5 - 2, 0, 26 . = 0 and 4 # 02 = 0 = 16 and 4 # 22 = 16 - 2 is a solution. 0 is a solution. 2 is a solution. r SECTION 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations EXAMPL E 8 117 Solving an Equation by Factoring Solve the equation: x3 - x2 - 4x + 4 = 0 Solution Do you recall the method of factoring by grouping? (If not, review pp. 53–54.) Group the terms of x3 - x2 - 4x + 4 = 0 as follows: 1x3 - x2 2 - 14x - 42 = 0 Factor out x2 from the first grouping and 4 from the second. x2 1x - 12 - 41x - 12 = 0 This reveals the common factor 1x - 12, so we have 1x2 1x - 22 1x x - 2 = x = - 42 1x - 12 + 22 1x - 12 0 or x + 2 2 x = = = = 0 0 Factor again. 0 or x - 1 = 0 Set each factor equal to 0. -2 x = 1 Solve. Check: x = - 2: 1 - 22 3 - 1 - 22 2 - 41 - 22 + 4 = - 8 - 4 + 8 + 4 = 0 - 2 is a solution. x = 1: 1 - 1 - 4112 + 4 = 1 - 1 - 4 + 4 = 0 1 is a solution. x = 2: 2 - 2 - 4122 + 4 = 8 - 4 - 8 + 4 = 0 2 is a solution. 3 3 2 2 The solution set is 5 - 2, 1, 26 . Now Work PROBLEM r 81 1.4 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. True or False The principal square root of any nonnegative real number is always nonnegative. (pp. 23–24) 3 2. 2 -8 = (pp. 73–77) 3. Factor 6x3 - 2x2 (pp. 49–55) Concepts and Vocabulary 4. True or False Factoring can only be used to solve quadratic equations or equations that are quadratic in form. 5. If u is an expression that involves x, then the equation au2 + bu + c = 0, a ≠ 0, is called an equation . 6. True or False Radical equations sometimes have extraneous solutions. 7. An apparent solution that does not satisfy the original equation is called a(n) solution. (a) extraneous (b) imaginary (c) radical (d) conditional 8. Which equation is likely to require squaring each side more than once? (a) 2x + 2 = 23x - 5 (b) x4 - 3x2 = 10 (c) 2x + 1 + 2x - 4 = 8 (d) 23x + 1 = 5 Skill Building In Problems 9–42, find the real solutions of each equation. 9. 22t - 1 = 1 10. 23t + 4 = 2 11. 23t + 4 = - 6 12. 25t + 3 = - 2 3 13. 2 1 - 2x - 3 = 0 3 14. 2 1 - 2x - 1 = 0 5 2 15. 2 x + 2x = - 1 4 2 16. 2 x + 16 = 25 17. x = 82x 18. x = 31x 19. 215 - 2x = x 20. 212 - x = x 21. x = 22x - 1 22. x = 22- x - 1 23. 2x2 - x - 4 = x + 2 24. 23 - x + x2 = x - 2 25. 3 + 23x + 1 = x 26. 2 + 212 - 2x = x 27. 23(x + 10) - 4 = x 28. 21 - x - 3 = x + 2 29. 22x + 3 - 2x + 1 = 1 30. 23x + 7 + 2x + 2 = 1 31. 23x + 1 - 2x - 1 = 2 32. 23x - 5 - 2x + 7 = 2 118 CHAPTER 1 Equations and Inequalities 33. 23 - 21x = 1x 37. 15x - 22 1>3 = 2 36. 13x - 52 1>2 = 2 39. 1x2 + 92 1>2 = 5 35. 13x + 12 1>2 = 4 34. 210 + 31x = 1x 40. 1x2 - 162 1>2 38. 12x + 12 1>3 = - 1 41. x3>2 - 3x1>2 = 0 = 9 42. x3>4 - 9x1>4 = 0 In Problems 43–74, find the real solutions of each equation. 43. x4 - 5x2 + 4 = 0 44. x4 - 10x2 + 25 = 0 45. 3x4 - 2x2 - 1 = 0 46. 2x4 - 5x2 - 12 = 0 47. x6 + 7x3 - 8 = 0 48. x6 - 7x3 - 8 = 0 49. 1x + 22 + 71x + 22 + 12 = 0 50. 12x + 52 - 12x + 52 - 6 = 0 51. 13x + 42 2 - 613x + 42 + 9 = 0 52. 12 - x2 2 + 12 - x2 - 20 = 0 53. 21s + 12 2 - 51s + 12 = 3 54. 311 - y2 2 + 511 - y2 + 2 = 0 55. x - 4x1x = 0 56. x + 81x = 0 57. x + 1x = 20 2 2 58. x + 1x = 6 59. t 61. 4x1>2 - 9x1>4 + 4 = 0 62. x1>2 - 3x1>4 + 2 = 0 4 63. 25x2 - 6 = x 4 64. 2 4 - 5x2 = x 65. x2 + 3x + 2x2 + 3x = 6 66. x2 - 3x - 2x2 - 3x = 2 67. 1 1 = + 2 x + 1 1x + 12 2 68. - 2t 1>4 + 1 = 0 60. z1>2 - 4z1>4 + 4 = 0 1 1 + = 12 x - 1 1x - 12 2 69. 3x -2 - 7x -1 - 6 = 0 72. 3x4>3 + 5x2>3 - 2 = 0 71. 2x2>3 - 5x1>3 - 3 = 0 70. 2x -2 - 3x -1 - 4 = 0 73. a 1>2 2 2v v b + = 8 v + 1 v + 1 74. a 2 y y b = 6a b + 7 y - 1 y - 1 In Problems 75–90, find the real solutions of each equation by factoring. 75. x3 - 9x = 0 77. 4x3 = 3x2 76. x4 - x2 = 0 78. x5 = 4x3 79. x3 + x2 - 20x = 0 80. x3 + 6x2 - 7x = 0 81. x3 + x2 - x - 1 = 0 82. x3 + 4x2 - x - 4 = 0 83. x3 - 3x2 - 4x + 12 = 0 84. x3 - 3x2 - x + 3 = 0 85. 2x3 + 4 = x2 + 8x 86. 3x3 + 4x2 = 27x + 36 87. 5x3 + 45x = 2x2 + 18 88. 3x3 + 12x = 5x2 + 20 89. x1x2 - 3x2 1>3 + 21x2 - 3x2 4>3 = 0 90. 3x1x2 + 2x2 1>2 - 21x2 + 2x2 3>2 = 0 In Problems 91–96, find the real solutions of each equation. Use a calculator to express any solutions rounded to two decimal places. 91. x - 4x1>2 + 2 = 0 92. x2>3 + 4x1>3 + 2 = 0 93. x4 + 23 x2 - 3 = 0 94. x4 + 22 x2 - 2 = 0 95. p11 + t2 2 = p + 1 + t 96. p11 + r2 2 = 2 + p11 + r2 Mixed Practice 97. If k = x + 3 and k 2 - k = 12, find x. x - 3 98. If k = x + 3 and k 2 - 3k = 28, find x. x - 4 Applications 99. Physics: Using Sound to Measure Distance The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If t 1 is the time (measured in seconds) that it takes for the object to strike the water, then t 1 will obey the equation s = 16t 21 , where s is the distance (measured in feet). It 1s follows that t 1 = . Suppose that t 2 is the time that it takes for the sound 4 of the impact to reach your ears. Because sound waves are known to travel at a speed of approximately 1100 feet per second, the time t 2 to travel the s . See the illustration. distance s will be t 2 = 1100 Sound waves: s t2 –––– 1100 Falling object: s t1 –– 4 SECTION 1.5 Solving Inequalities 119 Now t 1 + t 2 is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation Total time elapsed = s 1s + 4 1100 Find the distance to the water’s surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds. 100. Crushing Load A civil engineer relates the thickness T, in inches, and height H, in feet, of a square wooden pillar to its crushing 2 4 LH . If a square wooden pillar is 4 inches thick and 10 feet high, what is its crushing load? load L, in tons, using the model T = A 25 101. Foucault’s Pendulum The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The l period T, in seconds, is given by the formula T = 2p , where l is the length, in feet, of the pendulum. In 1851, Jean Bernard A 32 Leon Foucault demonstrated the axial rotation of Earth using a large pendulum that he hung in the Panthéon in Paris. The period of Foucault’s pendulum was approximately 16.5 seconds. What was its length? Explaining Concepts: Discussion and Writing 102. Make up a radical equation that has no solution. 103. Make up a radical equation that has an extraneous solution. 104. Discuss the step in the solving process for radical equations that leads to the possibility of extraneous solutions. Why is there no such possibility for linear and quadratic equations? 105. What Went Wrong? On an exam, Jane solved the equation 22x + 3 - x = 0 and wrote that the solution set was { - 1, 3}. Jane received 3 out of 5 points for the problem. Jane asks you why she received 3 out of 5 points. Provide an explanation. ‘Are You Prepared?’ Answers 1. True 2. - 2 3. 2x2 13x - 12 1.5 Solving Inequalities PREPARING FOR THIS SECTION Before getting started, review the following: r Algebra Essentials (Section R.2, pp. 17–26) Now Work the ‘Are You Prepared?’ problems on page 127. OBJECTIVES 1 2 3 4 Use Interval Notation (p. 120) Use Properties of Inequalities (p. 121) Solve Inequalities (p. 123) Solve Combined Inequalities (p. 124) Suppose that a and b are two real numbers and a 6 b. The notation a 6 x 6 b means that x is a number between a and b. The expression a 6 x 6 b is equivalent to the two inequalities a 6 x and x 6 b. Similarly, the expression a … x … b is equivalent to the two inequalities a … x and x … b. The remaining two possibilities, a … x 6 b and a 6 x … b, are defined similarly. Although it is acceptable to write 3 Ú x Ú 2, it is preferable to reverse the inequality symbols and write instead 2 … x … 3 so that the values go from smaller to larger, reading from left to right. A statement such as 2 … x … 1 is false because there is no number x for which 2 … x and x … 1. Finally, never mix inequality symbols, as in 2 … x Ú 3. 120 CHAPTER 1 Equations and Inequalities 1 Use Interval Notation Let a and b represent two real numbers with a 6 b. DEFINITION In Words The notation [a, b] represents all real numbers between a and b, inclusive. The notation (a, b) represents all real numbers between a and b, not including either a or b. An open interval, denoted by (a, b), consists of all real numbers x for which a 6 x 6 b. A closed interval, denoted by [a, b], consists of all real numbers x for which a … x … b. The half-open, or half-closed, intervals are (a, b], consisting of all real numbers x for which a 6 x … b, and [a, b), consisting of all real numbers x for which a … x 6 b. In each of these definitions, a is called the left endpoint and b the right endpoint of the interval. The symbol q (read as “infinity”) is not a real number, but notation used to indicate unboundedness in the positive direction. The symbol - q (read as “negative infinity”) also is not a real number, but notation used to indicate unboundedness in the negative direction. The symbols q and - q are used to define five other kinds of intervals: [a, ˆ ) (a, ˆ ) (− ˆ , a] (− ˆ , a) (− ˆ , ˆ ) Consists of all real numbers x for which x Consists of all real numbers x for which x Consists of all real numbers x for which x Consists of all real numbers x for which x Consists of all real numbers Ú 7 … 6 a a a a Note that q and - q are never included as endpoints, since neither is a real number. Table 1 summarizes interval notation, corresponding inequality notation, and their graphs. Table 1 EX AMPLE 1 Interval Inequality Graph The open interval (a, b) a<x<b The closed interval [a, b] a#x#b The half-open interval [a, b) a#x<b The half-open interval (a, b] a<x#b The interval [a, ∞) x$a The interval (a, ∞) x>a The interval (−∞, a] x#a The interval (−∞, a) x<a The interval (−∞, ∞) All real numbers a b a b a b a b a a a a Writing Inequalities Using Interval Notation Write each inequality using interval notation. (a) 1 … x … 3 Solution (b) - 4 6 x 6 0 (c) x 7 5 (d) x … 1 (a) 1 … x … 3 describes all real numbers x between 1 and 3, inclusive. In interval notation, we write 3 1, 34 . (b) In interval notation, - 4 6 x 6 0 is written 1 - 4, 02. SECTION 1.5 Solving Inequalities (c) In interval notation, x 7 5 is written 15, q 2. (d) In interval notation, x … 1 is written 1 - q , 14 . EXAMPL E 2 r Writing Intervals Using Inequality Notation Write each interval as an inequality involving x. (a) 3 1, 42 Solution 121 (a) (b) (c) (d) (b) 12, q 2 (c) 3 2, 34 (d) 1 - q , - 34 3 1, 42 consists of all real numbers x for which 1 … x 6 4. 12, q 2 consists of all real numbers x for which x 7 2. 3 2, 34 consists of all real numbers x for which 2 … x … 3. 1 - q , - 34 consists of all real numbers x for which x … - 3. Now Work PROBLEMS 13, 25, AND r 33 2 Use Properties of Inequalities The product of two positive real numbers is positive, the product of two negative real numbers is positive, and the product of 0 and 0 is 0. For any real number a, the value of a2 is 0 or positive; that is, a2 is nonnegative. This is called the nonnegative property. Nonnegative Property In Words The square of a real number is never negative. For any real number a, a2 Ú 0 (1) When the same number is added to both sides of an inequality, an equivalent inequality is obtained. For example, since 3 6 5, then 3 + 4 6 5 + 4 or 7 6 9. This is called the addition property of inequalities. In Words The addition property states that the sense, or direction, of an inequality remains unchanged if the same number is added to each side. Addition Property of Inequalities For real numbers a, b, and c, If a 6 b, then a + c 6 b + c. If a 7 b, then a + c 7 b + c. (2a) (2b) Figure 2 illustrates the addition property (2a). In Figure 2(a), we see that a lies to the left of b. If c is positive, then a + c and b + c lie c units to the right of a and c units to the right of b, respectively. Consequently, a + c must lie to the left of b + c; that is, a + c 6 b + c. Figure 2(b) illustrates the situation if c is negative. – c units c units c units a b a +c – c units b +c (a) If a < b and c > 0, then a + c < b + c. a+c b +c a b (b) If a < b and c < 0, then a + c < b + c. Figure 2 Addition property of inequalities D R AW A N I L LU S T R AT I O N S I M I L A R TO F I G U R E T H AT I L LU S T R AT E S T H E A D D I T I O N P R O P E R T Y 2 (2b). 122 CHAPTER 1 Equations and Inequalities EX AMPLE 3 Addition Property of Inequalities (a) If x 6 - 5, then x + 5 6 - 5 + 5 or x + 5 6 0. (b) If x 7 2, then x + 1 - 22 7 2 + 1 - 22 or x - 2 7 0. Now Work EX AMPLE 4 PROBLEM r 41 Multiplying an Inequality by a Positive Number Express as an inequality the result of multiplying each side of the inequality 3 6 7 by 2. Solution Begin with 3 6 7 Multiplying each side by 2 yields the numbers 6 and 14, so we have r 6 6 14 EX AMPLE 5 Multiplying an Inequality by a Negative Number Express as an inequality the result of multiplying each side of the inequality 9 7 2 by - 4. Solution Begin with 9 7 2 Multiplying each side by - 4 yields the numbers - 36 and - 8, so we have r - 36 6 - 8 In Words Multiplying by a negative number reverses the inequality. Note that the effect of multiplying both sides of 9 7 2 by the negative number - 4 is that the direction of the inequality symbol is reversed. Examples 4 and 5 illustrate the following general multiplication properties for inequalities: Multiplication Properties for Inequalities In Words The multiplication properties state that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive real number, whereas the direction is reversed if each side is multiplied by a negative real number. EX A MPLE 6 For real numbers a, b, and c, If a 6 b and if c 7 0, then ac 6 bc. If a 6 b and if c 6 0, then ac 7 bc. (3a) If a 7 b and if c 7 0, then ac 7 bc. If a 7 b and if c 6 0, then ac 6 bc. (3b) Multiplication Property of Inequalities 1 1 12x2 6 162 or x 6 3. 2 2 x x (b) If 7 12, then - 3a b 6 - 31122 or x 6 - 36. -3 -3 - 4x -8 (c) If - 4x 6 - 8, then 7 or x 7 2. -4 -4 (d) If - x 7 8, then 1 - 12 1 - x2 6 1 - 12 182 or x 6 - 8. (a) If 2x 6 6, then Now Work PROBLEM 47 r SECTION 1.5 Solving Inequalities In Words The reciprocal property states that the reciprocal of a positive real number is positive and that the reciprocal of a negative real number is negative. 123 Reciprocal Property for Inequalities 1 7 0. a 1 If a 6 0, then 6 0. a If a 7 0, then 1 7 0, then a 7 0. a 1 If 6 0, then a 6 0. a If (4a) (4b) 3 Solve Inequalities An inequality in one variable is a statement involving two expressions, at least one containing the variable, separated by one of the inequality symbols: 6 , … , 7 , or Ú . To solve an inequality means to find all values of the variable for which the statement is true. These values are called solutions of the inequality. For example, the following are all inequalities involving one variable x: x + 5 6 8 2x - 3 Ú 4 x2 - 1 … 3 x + 1 7 0 x - 2 As with equations, one method for solving an inequality is to replace it by a series of equivalent inequalities until an inequality with an obvious solution, such as x 6 3, is obtained. Equivalent inequalities are obtained by applying some of the same properties that are used to find equivalent equations. The addition property and the multiplication properties for inequalities form the basis for the following procedures. Procedures That Leave the Inequality Symbol Unchanged 1. Simplify both sides of the inequality by combining like terms and eliminating parentheses: Replace by x + 2 + 6 7 2x + 51x + 12 x + 8 7 7x + 5 2. Add or subtract the same expression on both sides of the inequality: Replace by 3x - 5 6 4 13x - 52 + 5 6 4 + 5 3. Multiply or divide both sides of the inequality by the same positive expression: Replace 4x 7 16 by 4x 16 7 4 4 Procedures That Reverse the Sense or Direction of the Inequality Symbol 1. Interchange the two sides of the inequality: Replace 3 6 x by x 7 3 2. Multiply or divide both sides of the inequality by the same negative expression: Replace - 2x 7 6 by - 2x 6 6 -2 -2 As the examples that follow illustrate, we solve inequalities using many of the same steps that we would use to solve equations. In writing the solution of an 124 CHAPTER 1 Equations and Inequalities inequality, either set notation or interval notation may be used, whichever is more convenient. EX AMPLE 7 Solving an Inequality Solve the inequality 3 - 2x 6 5, and graph the solution set. Solution 3 - 2x 6 5 3 - 2x - 3 6 5 - 3 Subtract 3 from both sides. - 2x 6 2 Simplify. - 2x 2 7 -2 -2 Divide both sides by - 2. (The sense of the inequality symbol is reversed.) x 7 -1 3 2 1 0 1 2 Simplify. The solution set is 5 x x 7 - 16 or, using interval notation, all numbers in the interval 1 - 1, q 2. See Figure 3 for the graph. r Figure 3 x 7 - 1 EX AMPLE 8 Solving an Inequality Solve the inequality 4x + 7 Ú 2x - 3, and graph the solution set. Solution 4x + 7 Ú 2x - 3 4x + 7 - 7 Ú 2x - 3 - 7 Subtract 7 from both sides. 4x Ú 2x - 10 Simplify. 4x - 2x Ú 2x - 10 - 2x 2x Ú - 10 Simplify. 2x - 10 Ú 2 2 Divide both sides by 2. (The direction of the inequality symbol is unchanged.) x Ú -5 6 5 4 3 2 1 Subtract 2x from both sides. Simplify. The solution set is 5 x x Ú - 56 or, using interval notation, all numbers in the interval 3 - 5, q 2. See Figure 4 for the graph. r Figure 4 x Ú - 5 Now Work PROBLEM 55 4 Solve Combined Inequalities EX AMPLE 9 Solving a Combined Inequality Solve the inequality - 5 6 3x - 2 6 1, and graph the solution set. Solution Recall that the inequality - 5 6 3x - 2 6 1 is equivalent to the two inequalities - 5 6 3x - 2 and 3x - 2 6 1 SECTION 1.5 Solving Inequalities 125 Solve each of these inequalities separately. -5 -5 + 2 -3 -3 3 -1 6 3x - 2 6 3x - 2 + 2 6 3x 3x 6 3 6 x Add 2 to both sides. Simplify. Divide both sides by 3. Simplify. 3x - 2 6 1 3x - 2 + 2 6 1 + 2 3x 6 3 3x 3 6 3 3 x 6 1 The solution set of the original pair of inequalities consists of all x for which - 1 6 x and x 6 1 3 2 1 0 1 This may be written more compactly as 5 x - 1 6 x 6 16 . In interval notation, the solution is 1 - 1, 12. See Figure 5 for the graph. 2 r Figure 5 - 1 6 x 6 1 Observe in the preceding process that solving each of the two inequalities required exactly the same steps. A shortcut to solving the original inequality algebraically is to deal with the two inequalities at the same time, as follows: -5 -5 + 2 -3 -3 3 -1 6 3x - 2 6 1 6 3x - 2 + 2 6 1 + 2 Add 2 to each part. 6 3x 6 3 Simplify. 3x 3 6 6 Divide each part by 3. 3 3 Simplify. 6 x 6 1 Solving a Combined Inequality EX AM PL E 1 0 Solve the inequality - 1 … Solution -1 … 3 - 5x 2 … 9 3 - 5x b … 2(9) 2 Multiply each part by 2 to remove the denominator. 3 - 5x Simplify. 2( - 12 … 2a -2 … 3 - 5x … 9, and graph the solution set. 2 … 18 - 2 - 3 … 3 - 5x - 3 … 18 - 3 Subtract 3 from each part to isolate 4 3 2 1 Figure 6 - 3 … x … 1 0 1 2 -5 … - 5x … 15 -5 Ú -5 - 5x -5 Ú the term containing x. Simplify. 15 -5 Divide each part by - 5 (reverse the direction of each inequality symbol). 1 Ú x Ú -3 Simplify. -3 … x … 1 Reverse the order so that the numbers get larger as you read from left to right. The solution set is 5 x - 3 … x … 16 , that is, all x in the interval 3 - 3, 14 . Figure 6 illustrates the graph. r Now Work PROBLEM 75 126 CHAPTER 1 Equations and Inequalities EX A MPL E 11 Solution Using the Reciprocal Property to Solve an Inequality Solve the inequality 14x - 12 -1 7 0, and graph the solution set. 1 Recall that 14x - 12 -1 = . The Reciprocal Property states that when 4x - 1 1 7 0, then a 7 0. a 14x - 12 -1 7 0 1 7 0 4x - 1 4x - 1 7 0 Reciprocal Property 4x 7 1 Add 1 to both sides. x 7 0 1 – 4 1 1 4 Divide both sides by 4. 1 1 The solution set is e x 0 x 7 f, that is, all x in the interval a , q b . Figure 7 4 4 illustrates the graph. r 1 Figure 7 x 7 4 Now Work EX A MPL E 1 2 PROBLEM 85 Creating Equivalent Inequalities If - 1 6 x 6 4, find a and b so that a 6 2x + 1 6 b. Solution The idea here is to change the middle part of the combined inequality from x to 2x + 1, using properties of inequalities. -1 6 x 6 4 - 2 6 2x 6 8 - 1 6 2x + 1 6 9 Multiply each part by 2. Add 1 to each part. r Now we see that a = - 1 and b = 9. Now Work PROBLEM 95 Application EX A MPL E 1 3 Physics: Ohm’s Law In electricity, Ohm’s law states that E = IR, where E is the voltage (in volts), I is the current (in amperes), and R is the resistance (in ohms). An air-conditioning unit is rated at a resistance of 10 ohms. If the voltage varies from 110 to 120 volts, inclusive, what corresponding range of current will the air conditioner draw? Solution The voltage lies between 110 and 120, inclusive, so 110 110 110 110 10 11 … E … IR … I 1102 I 1102 … 10 … I … 120 … 120 … 120 120 … 10 … 12 Ohm’s law, E = IR R = 10 Divide each part by 10. Simplify. The air conditioner will draw between 11 and 12 amperes of current, inclusive. r SECTION 1.5 Solving Inequalities 127 1.5 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Graph the inequality: x Ú - 2. (pp. 17–26) 2. True or False - 5 7 - 3 (pp. 17–26) Concepts and Vocabulary 3. A(n) , denoted 3a, b4, consists of all real numbers x for which a … x … b. 9. True or False The square of any real number is always nonnegative. 10. True or False A half-closed interval must have an endpoint of either - q or q . state that the sense, 4. The or direction, of an inequality remains the same if each side is multiplied by a positive number, while the direction is reversed if each side is multiplied by a negative number. 11. Which of the following will change the direction, or sense, of an inequality? (a) Dividing each side by a positive number (b) Interchanging sides (c) Adding a negative number to each side (d) Subtracting a positive number from each side In Problems 5–8, assume that a 6 b and c 6 0. 5. True or False a + c 6 b + c 6. True or False a - c 6 b - c 7. True or False ac 7 bc 12. Which pair of inequalities is equivalent to 0 6 x … 3? (a) x 7 0 and x Ú 3 (b) x 6 0 and x Ú 3 (c) x 7 0 and x … 3 (d) x 6 0 and x … 3 b a 6 8. True or False c c Skill Building In Problems 13–18, express the graph shown in blue using interval notation. Also express each as an inequality involving x. 13. 14. –1 0 1 2 3 –2 –1 0 1 2 16. 15. –2 –1 0 1 2 –1 0 1 2 3 17. –1 0 1 2 3 –1 0 1 2 3 18. In Problems 19–24, an inequality is given. Write the inequality obtained by: (a) Adding 3 to each side of the given inequality. (b) Subtracting 5 from each side of the given inequality. (c) Multiplying each side of the given inequality by 3. (d) Multiplying each side of the given inequality by - 2. 19. 3 6 5 21. 4 7 - 3 20. 2 7 1 22. - 3 7 - 5 23. 2x + 1 6 2 24. 1 - 2x 7 5 In Problems 25–32, write each inequality using interval notation, and graph each inequality on the real number line. 25. 0 … x … 4 26. - 1 6 x 6 5 27. 4 … x 6 6 28. - 2 6 x 6 0 29. x Ú 4 30. x … 5 31. x 6 - 4 32. x 7 1 In Problems 33–40, write each interval as an inequality involving x, and graph each inequality on the real number line. 33. 32, 54 34. 11, 22 35. 1 - 3, - 22 36. 30, 12 37. 34, q 2 38. 1 - q , 24 39. 1 - q , - 32 40. 1 - 8, q 2 In Problems 41–54, fill in the blank with the correct inequality symbol. 41. If x 6 5, then x - 5 42. If x 6 - 4, then x + 4 0. 43. If x 7 - 4, then x + 4 0. 0. 44. If x 7 6, then x - 6 0. 45. If x Ú - 4, then 3x - 12. 46. If x … 3, then 2x 47. If x 7 6, then - 2x - 12. 48. If x 7 - 2, then - 4x 8. 49. If x Ú 5, then - 4x - 20. 50. If x … - 4, then - 3x 12. 51. If 2x 7 6, then x 53. If - 1 x … 3, then x 2 52. If 3x … 12, then x 3. - 6. 54. If - 1 x 7 1, then x 4 6. 4. - 4. 128 CHAPTER 1 Equations and Inequalities In Problems 55–92, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. 55. x + 1 6 5 56. x - 6 6 1 57. 1 - 2x … 3 58. 2 - 3x … 5 59. 3x - 7 7 2 60. 2x + 5 7 1 61. 3x - 1 Ú 3 + x 62. 2x - 2 Ú 3 + x 63. - 21x + 32 6 8 64. - 311 - x2 6 12 65. 4 - 311 - x2 … 3 66. 8 - 412 - x2 … - 2x 67. 1 1x - 42 7 x + 8 2 68. 3x + 4 7 70. x x Ú 2 + 3 6 71. 0 … 2x - 6 … 4 72. 4 … 2x + 2 … 10 74. - 3 … 3 - 2x … 9 75. - 3 6 73. - 5 … 4 - 3x … 2 76. 0 6 3x + 2 6 4 2 77. 1 6 1 - 1 1x - 22 3 69. 1 x 6 4 2 80. 1x - 12 1x + 12 7 1x - 32 1x + 42 82. x19x - 52 … 13x - 12 2 83. 85. 14x + 22 -1 6 0 86. 12x - 12 -1 7 0 88. 2(3x + 5)-1 … - 3 89. 0 6 1 x + 1 3 … 6 2 3 4 1 x 6 1 3 81. x14x + 32 … 12x + 12 2 84. 1 x + 1 2 6 … 3 2 3 87. (2 - 7x)-1 Ú 5 3 2 6 x 5 90. 0 6 92. 0 6 13x + 62 -1 6 1 2 2x - 1 6 0 4 78. 0 6 1 - 79. 1x + 22 1x - 32 7 1x - 12 1x + 12 91. 0 6 12x - 42 -1 6 x x Ú 1 2 4 4 2 6 x 3 1 3 Applications and Extensions In Problems 93–102, find a and b. 93. If - 1 6 x 6 1, then a 6 x + 4 6 b. 94. If - 3 6 x 6 2, then a 6 x - 6 6 b. 95. If 2 6 x 6 3, then a 6 - 4x 6 b. 1 96. If - 4 6 x 6 0, then a 6 x 6 b. 2 97. If 0 6 x 6 4, then a 6 2x + 3 6 b. 98. If - 3 6 x 6 3, then a 6 1 - 2x 6 b. 99. If - 3 6 x 6 0, then a 6 100. If 2 6 x 6 4, then a 6 1 6 b. x + 4 30-year-old female in 2014 could expect to live at least 55.6 more years. (a) To what age could an average 30-year-old male expect to live? Express your answer as an inequality. (b) To what age could an average 30-year-old female expect to live? Express your answer as an inequality. (c) Who can expect to live longer, a male or a female? By how many years? Source: Social Security Administration, 2014 1 6 b. x - 6 101. If 6 6 3x 6 12, then a 6 x2 6 b. 102. If 0 6 2x 6 6, then a 6 x2 6 b. 103. What is the domain of the variable in the expression 13x + 6 ? 104. What is the domain of the variable in the expression 18 + 2x ? 105. A young adult may be defined as someone older than 21 but less than 30 years of age. Express this statement using inequalities. 106. Middle-aged may be defined as being 40 or more and less than 60. Express this statement using inequalities. 107. Life Expectancy The Social Security Administration determined that an average 30-year-old male in 2014 could expect to live at least 51.9 more years and that an average JAN 2014 JULY 2069 NOV 2065 108. General Chemistry For a certain ideal gas, the volume V (in cubic centimeters) equals 20 times the temperature T (in degrees Celsius). If the temperature varies from 80° to 120°C, inclusive, what is the corresponding range of the volume of the gas? 109. Real Estate A real estate agent agrees to sell an apartment complex according to the following commission schedule: $45,000 plus 25% of the selling price in excess of $900,000. Assuming that the complex will sell at some price between SECTION 1.5 Solving Inequalities $900,000 and $1,100,000, inclusive, over what range does the agent’s commission vary? How does the commission vary as a percent of selling price? 110. Sales Commission A used car salesperson is paid a commission of $25 plus 40% of the selling price in excess of owner’s cost. The owner claims that used cars typically sell for at least owner’s cost plus $200 and at most owner’s cost plus $3000. For each sale made, over what range can the salesperson expect the commission to vary? 111. Federal Tax Withholding The percentage method of withholding for federal income tax (2014) states that a single person whose weekly wages, after subtracting withholding allowances, are over $753, but not over $1762, shall have $97.75 plus 25% of the excess over $753 withheld. Over what range does the amount withheld vary if the weekly wages vary from $900 to $1100, inclusive? Source: Employer’s Tax Guide. Internal Revenue Service, 2014. 112. Exercising Sue wants to lose weight. For healthy weight loss, the American College of Sports Medicine (ACSM) recommends 200 to 300 minutes of exercise per week. For the first six days of the week, Sue exercised 40, 45, 0, 50, 25, and 35 minutes. How long should Sue exercise on the seventh day in order to stay within the ACSM guidelines? 113. Electricity Rates Commonwealth Edison Company’s charge for electricity in January 2014 was 8.21¢ per kilowatt-hour. In addition, each monthly bill contains a customer charge of $15.37. If last year’s bills ranged from a low of $72.84 to a high of $237.04, over what range did usage vary (in kilowatt-hours)? Source: Commonwealth Edison Co., 2014. 114. Water Bills The Village of Oak Lawn charges homeowners $57.07 per quarter-year plus $5.81 per 1000 gallons for water usage in excess of 10,000 gallons. In 2014 one homeowner’s quarterly bill ranged from a high of $150.03 to a low of $97.74. Over what range did water usage vary? Source: Village of Oak Lawn, Illinois, January 2014. 115. Markup of a New Car The markup over dealer’s cost of a new car ranges from 12% to 18%. If the sticker price is $18,000, over what range will the dealer’s cost vary? 116. IQ Tests A standard intelligence test has an average score of 100. According to statistical theory, of the people who take the test, the 2.5% with the highest scores will have scores of more than 1.96s above the average, where s (sigma, a number called the standard deviation) depends on the nature of the test. If s = 12 for this test and there is (in principle) no upper limit to the score possible on the test, write the interval of possible test scores of the people in the top 2.5%. 129 117. Computing Grades In your Economics 101 class, you have scores of 68, 82, 87, and 89 on the first four of five tests. To get a grade of B, the average of the first five test scores must be greater than or equal to 80 and less than 90. (a) Solve an inequality to find the least score you can get on the last test and still earn a B. (b) What score do you need if the fifth test counts double? What do I need to get a B? 82 68 89 87 118. “Light” Foods For food products to be labeled “light,” the U.S. Food and Drug Administration requires that the altered product must either contain at least one-third fewer calories than the regular product or contain at least one-half less fat than the regular product. If a serving of Miracle Whip® Light contains 20 calories and 1.5 grams of fat, then what must be true about either the number of calories or the grams of fat in a serving of regular Miracle Whip®? a + b 6 b. The 119. Arithmetic Mean If a 6 b, show that a 6 2 a+b number is called the arithmetic mean of a and b. 2 120. Refer to Problem 119. Show that the arithmetic mean of a and b is equidistant from a and b. 121. Geometric Mean If 0 6 a 6 b, show that a 6 1ab 6 b. The number 1ab is called the geometric mean of a and b. 122. Refer to Problems 119 and 121. Show that the geometric mean of a and b is less than the arithmetic mean of a and b. 123. Harmonic Mean For 0 6 a 6 b, let h be defined by 1 1 1 1 = a + b h 2 a b Show that a 6 h 6 b. The number h is called the harmonic mean of a and b. 124. Refer to Problems 119, 121, and 123. Show that the harmonic mean of a and b equals the geometric mean squared, divided by the arithmetic mean. 125. Another Reciprocal Property Prove that if 0 6 a 6 b, then 1 1 6 . 0 6 b a Explaining Concepts: Discussion and Writing 126. Make up an inequality that has no solution. Make up an inequality that has exactly one solution. 127. The inequality x2 + 1 6 - 5 has no real solution. Explain why. 128. Do you prefer to use inequality notation or interval notation to express the solution to an inequality? Give your reasons. Are there particular circumstances when you prefer one to the other? Cite examples. ‘Are You Prepared?’ Answers 1. 4 2 2. False 0 129. How would you explain to a fellow student the underlying reason for the multiplication properties for inequalities (page 122)? That is, the sense or direction of an inequality remains the same if each side is multiplied by a positive real number, whereas the direction is reversed if each side is multiplied by a negative real number. 130 CHAPTER 1 Equations and Inequalities 1.6 Equations and Inequalities Involving Absolute Value PREPARING FOR THIS SECTION Before getting started, review the following: r Algebra Essentials (Chapter R, Section R.2, pp. 17–26) Now Work the ‘Are You Prepared?’ problems on page 132. OBJECTIVES 1 Solve Equations Involving Absolute Value (p. 130) 2 Solve Inequalities Involving Absolute Value (p. 130) 1 Solve Equations Involving Absolute Value Recall that on the real number line, the absolute value of a equals the distance from the origin to the point whose coordinate is a. For example, there are two points whose distance from the origin is 5 units, - 5 and 5. So the equation 0 x 0 = 5 will have the solution set 5 - 5, 56 . This leads to the following result: THEOREM If a is a positive real number and if u is any algebraic expression, then 0 u 0 = a is equivalent to u = a or u = - a (1) Equation (1) requires that a be a positive number. If a = 0, equation (1) becomes |u| = 0 which is equivalent to u = 0. If a is less than zero, the equation has no real solution. To see why, consider the equation |x| = - 2. Because there is no real number whose distance from 0 on the real number line is negative, this equation has no real solution. EX AMPLE 1 Solving an Equation Involving Absolute Value Solve the equations: (a) 0 x + 4 0 = 13 Solution (b) 0 2x - 3 0 + 2 = 7 (a) This follows the form of equation (1), where u = x + 4. There are two possibilities: x + 4 = 13 or x + 4 = - 13 x = 9 or x = - 17 The solution set is 5 - 17, 96 . (b) The equation 0 2x - 3 0 + 2 = 7 is not in the form of equation (1). Proceed as follows: 0 2x - 3 0 + 2 = 7 0 2x - 3 0 = 5 2x - 3 = 5 2x = 8 x = 4 or or or Subtract 2 from each side. 2x - 3 = - 5 Apply (1). 2x = - 2 Add 3 to both sides. x = - 1 Divide both sides by 2. The solution set is 5 - 1, 46 . Now Work PROBLEM r 11 2 Solve Inequalities Involving Absolute Value EX AMPLE 2 Solving an Inequality Involving Absolute Value Solve the inequality: 0 x 0 6 4 SECTION 1.6 Equations and Inequalities Involving Absolute Value Solution Less than 4 units from origin O 0 1 2 We are looking for all points whose coordinate x is a distance less than 4 units from the origin. See Figure 8 for an illustration. Because any number x between - 4 and 4 satisfies the condition 0 x 0 6 4, the solution set consists of all numbers x for which - 4 6 x 6 4, that is, all x in the interval 1 - 4, 42. r O 25 24 23 22 21 131 3 4 Figure 8 x 6 4 THEOREM If a is a positive number and if u is an algebraic expression, then 0 u 0 6 a is equivalent to - a 6 u 6 a 0 u 0 … a is equivalent to - a … u … a (2) (3) In other words, 0 u 0 6 a is equivalent to - a 6 u and u 6 a. –a 0 a Figure 9 u … a, a 7 0 See Figure 9 for an illustration of statement (3). EXAMPL E 3 Solving an Inequality Involving Absolute Value Solve the inequality 0 2x + 4 0 … 3, and graph the solution set. 0 2x + 4 0 … 3 Solution -3 -3 - 4 -7 -7 2 7 2 5 7 –2 2 1 –2 0 2 4 EXAMPL E 4 1 3– 2 Figure 11 1 - 4x 6 5 Subtract 4 from each part. Simplify. Divide each part by 2. Simplify. 7 1 7 1 … x … - f, that is, all x in the interval c - , - d . See 2 2 2 2 Figure 10 for the graph of the solution set. Solving an Inequality Involving Absolute Value Solve the inequality 0 1 - 4x 0 6 5, and graph the solution set. 0 1 - 4x 0 6 5 -5 -5 - 1 -6 -6 -4 3 2 2 Apply statement (3). The solution set is e x ` - Solution 0 … 2x + 4 … 3 … 2x + 4 - 4 … 3 - 4 … 2x … -1 2x -1 … … 2 2 1 x … … 2 r Figure 10 2x + 4 … 3 5 4 3 2 1 This follows the form of statement (3); the expression u = 2x + 4 is inside the absolute value bars. 3 4 6 1 - 4x 6 5 6 1 - 4x - 1 6 5 - 1 6 - 4x 6 4 - 4x 4 7 7 -4 -4 7 x 7 -1 -1 6 x 6 3 2 This expression follows the form of statement (2); the expression u = 1 - 4x is inside the absolute value bars. Apply statement (2). Subtract 1 from each part. Simplify. Divide each part by - 4, which reverses the direction of the inequality symbols. Simplify. Rearrange the ordering. The solution set is e x 0 - 1 6 x 6 3 3 f, that is, all x in the interval a - 1, b . See 2 2 Figure 11 for the graph of the solution set. r Now Work PROBLEM 41 132 CHAPTER 1 Equations and Inequalities EX AMPLE 5 Solution 5 4 3 2 1 0 1 2 3 4 Solving an Inequality Involving Absolute Value Solve the inequality 0 x 0 7 3, and graph the solution set. We are looking for all points whose coordinate x is a distance greater than 3 units from the origin. Figure 12 illustrates the situation. Any number x less than - 3 or greater than 3 satisfies the condition 0 x 0 7 3. The solution set consists of all numbers x for which x 6 - 3 or x 7 3, that is, all x in 1 - q , - 32 h 13, q 2.* r Figure 12 x 7 3 THEOREM If a is a positive number and u is an algebraic expression, then 0 u 0 7 a is equivalent to u 6 - a or u 7 a (4) 0 u 0 Ú a is equivalent to u … - a or u Ú a –a 0 a Figure 13 u Ú a, a 7 0 See Figure 13 for an illustration of statement (5). Solving an Inequality Involving Absolute Value EX AMPLE 6 Solve the inequality 0 2x - 5 0 7 3, and graph the solution set. 0 2x - 5 0 7 3 Solution 2x - 5 2x - 5 + 5 2x 2x 2 x 2 1 0 1 2 (5) 3 4 5 6 7 Figure 14 2x - 5 7 3 This follows the form of statement (4); the expression u = 2x - 5 is inside the absolute value bars. 6 -3 or 2x - 5 7 6 - 3 + 5 or 2x - 5 + 5 7 6 2 or 2x 7 2 2x 6 or 7 2 2 6 1 or x 7 3 Apply statement (4). 3 + 5 Add 5 to each part. 8 Simplify. 8 Divide each part by 2. 2 4 Simplify. The solution set is 5 x x 6 1 or x 7 46 , that is, all x in 1 - q , 12 h 14, q 2. See Figure 14 for the graph of the solution set. r WARNING A common error to be avoided is to attempt to write the solution x 6 1 or x 7 4 as the combined inequality 1 7 x 7 4, which is incorrect, since there are no numbers x for which 1 7 x and x 7 4. Another common error is to “mix” the symbols and write 1 6 x 7 4, which makes no sense. ■ Now Work PROBLEM 45 *Recall that the symbol h stands for the union of two sets. Refer to page 2 if necessary. 1.6 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. 0 - 2 0 = 2. True or False 0 x 0 Ú 0 for any real number x. (p. 19) (p. 19) Concepts and Vocabulary 3. The solution set of the equation 0 x 0 = 5 is 5 4. The solution set of the inequality 0 x 0 6 5 is {x 0 }. 5. True or False The equation 0 x 0 = - 2 has no solution. 6. 6. True or False The inequality 0 x 0 Ú - 2 has the set of real numbers as its solution set. 7. Which of the following pairs of inequalities is equivalent to 0 x 0 7 4? (a) x 7 - 4 and x 6 4 (b) x 6 - 4 and x 6 4 (c) x 7 - 4 or x 7 4 (d) x 6 - 4 or x 7 4 8. Which of the following has no solution? (a) 0 x 0 6 - 5 (b) 0 x 0 … 0 (c) 0 x 0 7 0 (d) 0 x 0 Ú 0 SECTION 1.6 Equations and Inequalities Involving Absolute Value 133 Skill Building In Problems 9–36, find the real solutions, if any, of each equation. 9. 0 2x 0 = 6 13. 0 1 - 4t 0 + 8 = 13 10. 0 3x 0 = 12 14. 0 1 - 2z 0 + 6 = 9 11. 0 2x + 3 0 = 5 17. 0 - 2 0 x = 4 18. 0 3 0 x = 9 19. 22. ` 23. 0 u - 2 0 = - 21. ` 2 x + ` = 2 3 5 x 1 - ` = 1 2 3 12. 0 3x - 1 0 = 2 16. 0 - x 0 = 0 1 0 15. 0 - 2x 0 = 0 8 0 2 0x0 = 9 3 20. 3 0x0 = 9 4 24. 0 2 - v 0 = - 1 1 2 25. 4 - 0 2x 0 = 3 1 26. 5 - ` x ` = 3 2 27. 0 x2 - 9 0 = 0 28. 0 x2 - 16 0 = 0 29. 0 x2 - 2x 0 = 3 30. 0 x2 + x 0 = 12 31. 0 x2 + x - 1 0 = 1 32. 0 x2 + 3x - 2 0 = 2 33. ` 34. ` 35. 0 x2 + 3x 0 = 0 x2 - 2x 0 36. 0 x2 - 2x 0 = 0 x2 + 6x 0 3x - 2 ` = 2 2x - 3 2x + 1 ` = 1 3x + 4 In Problems 37–64, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. 37. 0 2x 0 6 8 38. 0 3x 0 6 15 39. 0 3x 0 7 12 40. 0 2x 0 7 6 41. 0 x - 2 0 + 2 6 3 42. 0 x + 4 0 + 3 6 5 43. 0 3t - 2 0 … 4 45. 0 2x - 3 0 Ú 2 46. 0 3x + 4 0 Ú 2 47. 0 1 - 4x 0 - 7 6 - 2 44. 0 2u + 5 0 … 7 49. 0 1 - 2x 0 7 3 50. 0 2 - 3x 0 7 1 51. 0 - 4x 0 + 0 - 5 0 … 1 52. 0 - x 0 - 0 4 0 … 2 53. 0 - 2x 0 7 0 - 3 0 54. 0 - x - 2 0 Ú 1 55. - 0 2x - 1 0 Ú - 3 56. - 0 1 - 2x 0 Ú - 3 57. 0 2x 0 6 - 1 58. 0 3x 0 Ú 0 59. 0 5x 0 Ú - 1 60. 0 6x 0 6 - 2 61. ` 62. 3 - 0 x + 1 0 6 2x + 3 1 - ` 6 1 3 2 1 2 63. 5 + 0 x - 1 0 7 1 2 48. 0 1 - 2x 0 - 4 6 - 1 64. ` 2x - 3 1 + ` 7 1 2 3 Applications and Extensions 65. Body Temperature “Normal” human body temperature is 98.6°F. If a temperature x that differs from normal by at least 1.5° is considered unhealthy, write the condition for an unhealthy temperature x as an inequality involving an absolute value, and solve for x. 66. Household Voltage In the United States, normal household voltage is 110 volts. However, it is not uncommon for actual voltage to differ from normal voltage by at most 5 volts. Express this situation as an inequality involving an absolute value. Use x as the actual voltage and solve for x. 67. Reading Books A HuffPost/YouGov poll conducted September 27–28, 2013, found that Americans read an average of 13.6 books per year. Suppose HuffPost/YouGov is 99% confident that the result from this poll is off by fewer than 1.8 books from the actual average x. Express this situation as an inequality involving absolute value, and solve the inequality for x to determine the interval in which the actual average is likely to fall. [Note: In statistics, this interval is called a 99% confidence interval.] 68. Speed of Sound According to data from the Hill Aerospace Museum (Hill Air Force Base, Utah), the speed of sound varies depending on altitude, barometric pressure, and temperature. For example, at 20,000 feet, 13.75 inches of mercury, and - 12.3°F, the speed of sound is about 707 miles per hour, but the speed can vary from this result by as much as 55 miles per hour as conditions change. (a) Express this situation as an inequality involving an absolute value. (b) Using x for the speed of sound, solve for x to find an interval for the speed of sound. 1 69. Express the fact that x differs from 3 by less than as an 2 inequality involving an absolute value. Solve for x. 70. Express the fact that x differs from - 4 by less than 1 as an inequality involving an absolute value. Solve for x. 71. Express the fact that x differs from - 3 by more than 2 as an inequality involving an absolute value. Solve for x. 72. Express the fact that x differs from 2 by more than 3 as an inequality involving an absolute value. Solve for x. 134 CHAPTER 1 Equations and Inequalities 82. Prove that 0 a - b 0 Ú 0 a 0 - 0 b 0 . In Problems 73–78, find a and b. 73. If 0 x - 1 0 6 3, then a 6 x + 4 6 b. [Hint: Apply the triangle inequality from Problem 81 to 0 a 0 = 0 1a - b2 + b 0 .] 74. If 0 x + 2 0 6 5, then a 6 x - 2 6 b. 75. If 0 x + 4 0 … 2, then a … 2x - 3 … b. 83. If a 7 0, show that the solution set of the inequality 76. If 0 x - 3 0 … 1, then a … 3x + 1 … b. 77. If 0 x - 2 0 … 7, then a … x2 6 a 1 … b. x - 10 consists of all numbers x for which - 1a 6 x 6 1a 1 … b. 78. If 0 x + 1 0 … 3, then a … x + 5 84. If a 7 0, show that the solution set of the inequality x2 7 a 79. Show that: if a 7 0, b 7 0, and 1a 6 1b, then a 6 b. [Hint: b - a = 1 1b - 1a2 1 1b + 1a2 .] consists of all numbers x for which 80. Show that a … 0 a 0 . x 6 - 1a or x 7 1a 81. Prove the triangle inequality 0 a + b 0 … 0 a 0 + 0 b 0 . [Hint: Expand 0 a + b 0 2 = 1a + b2 2, and use the result of Problem 80.] In Problems 85–92, use the results found in Problems 83 and 84 to solve each inequality. 85. x2 6 1 86. x2 6 4 87. x2 Ú 9 88. x2 Ú 1 89. x2 … 16 90. x2 … 9 91. x2 7 4 92. x2 7 16 93. Solve 0 3x - 0 2x + 1 0 0 = 4. 94. Solve 0 x + 0 3x - 2 0 0 = 2. Explaining Concepts: Discussion and Writing 95. The equation 0 x 0 = - 2 has no solution. Explain why. 96. The inequality 0 x 0 7 - 0.5 has all real numbers as solutions. Explain why. 97. The inequality 0 x 0 7 0 has as solution set 5 x x ≠ 06. Explain why. ‘Are You Prepared?’ Answers 1. 2 2. True 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications OBJECTIVES 1 2 3 4 5 Translate Verbal Descriptions into Mathematical Expressions (p. 135) Solve Interest Problems (p. 136) Solve Mixture Problems (p. 137) Solve Uniform Motion Problems (p. 138) Solve Constant Rate Job Problems (p. 140) Applied (word) problems do not come in the form “Solve the equation .c” Instead, they supply information using words, a verbal description of the real problem. So, to solve applied problems, we must be able to translate the verbal description into the language of mathematics. This can be done by using variables to represent unknown quantities and then finding relationships (such as equations) that involve these variables. The process of doing all this is called mathematical modeling. An equation or inequality that describes a relationship among the variables is called a model. SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 135 Any solution to the mathematical problem must be checked against the mathematical problem, the verbal description, and the real problem. See Figure 15 for an illustration of the modeling process. Real problem Verbal description Language of mathematics Mathematical problem Check Check Check Solution Figure 15 1 Translate Verbal Descriptions into Mathematical Expressions EXAMPL E 1 Translating Verbal Descriptions into Mathematical Expressions (a) For uniform motion, the average speed of an object equals the distance traveled divided by the time required. Translation: If r is the speed, d the distance, and t the time, then r = d . t (b) Let x denote a number. The number 5 times as large as x is 5x. The number 3 less than x is x - 3. The number that exceeds x by 4 is x + 4. The number that, when added to x, gives 5 is 5 - x. Now Work PROBLEM r 9 Always check the units used to measure the variables of an applied problem. In Example (1a), if r is measured in miles per hour, then the distance d must be expressed in miles, and the time t must be expressed in hours. It is a good practice to check units to be sure that they are consistent and make sense. The steps to follow for solving applied problems, given earlier, are repeated next. Steps for Solving Applied Problems STEP 1: STEP 2: STEP 3: STEP 4: STEP 5: Read the problem carefully, perhaps two or three times. Pay particular attention to the question being asked in order to identify what you are looking for. Identify any relevant formulas you may need (d = rt, A = pr 2, etc.). If you can, determine realistic possibilities for the answer. Assign a letter (variable) to represent what you are looking for, and if necessary, express any remaining unknown quantities in terms of this variable. Make a list of all the known facts, and translate them into mathematical expressions. These may take the form of an equation or an inequality involving the variable. If possible, draw an appropriately labeled diagram to assist you. Sometimes, creating a table or chart helps. Solve for the variable, and then answer the question. Check the answer with the facts in the problem. If it agrees, congratulations! If it does not agree, try again. 136 CHAPTER 1 Equations and Inequalities 2 Solve Interest Problems Interest is money paid for the use of money. The total amount borrowed (whether by an individual from a bank in the form of a loan, or by a bank from an individual in the form of a savings account) is called the principal. The rate of interest, expressed as a percent, is the amount charged for the use of the principal for a given period of time, usually on a yearly (that is, per annum) basis. Simple Interest Formula If a principal of P dollars is borrowed for a period of t years at a per annum interest rate r, expressed as a decimal, the interest I charged is I = Prt (1) Interest charged according to formula (1) is called simple interest. When using formula (1), be sure to express r as a decimal. For example, if the rate of interest is 4%, then r = 0.04. EX AMPLE 2 Finance: Computing Interest on a Loan Suppose that Juanita borrows $500 for 6 months at the simple interest rate of 9% per annum. What is the interest that Juanita will be charged on the loan? How much does Juanita owe after 6 months? Solution The rate of interest is given per annum, so the actual time that the money is borrowed must be expressed in years. The interest charged would be the principal, $500, times 1 the rate of interest (9, = 0.09), times the time in years, : 2 1 Interest charged = I = Prt = 15002 10.092 a b = +22.50 2 After 6 months, Juanita will owe what she borrowed plus the interest: r +500 + +22.50 = +522.50 EX AMPLE 3 Financial Planning Candy has $70,000 to invest and wants an annual return of $2800, which requires an overall rate of return of 4%. She can invest in a safe, government-insured certificate of deposit, but it pays only 2%. To obtain 4%, she agrees to invest some of her money in noninsured corporate bonds paying 7%. How much should be placed in each investment to achieve her goal? Solution STEP 1: The question is asking for two dollar amounts: the principal to invest in the corporate bonds and the principal to invest in the certificate of deposit. STEP 2: Let b represent the amount (in dollars) to be invested in the bonds. Then 70,000 - b is the amount that will be invested in the certificate. (Do you see why?) STEP 3: We set up a table: Bonds Certificate Total Principal ($) Rate Time (yr) b 7% = 0.07 1 0.07b Interest ($) 70,000 - b 2% = 0.02 1 0.02(70,000 - b) 70,000 4% = 0.04 1 0.04(70,000) = 2800 SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 137 Since the combined interest from the investments is equal to the total interest, we have Bond interest + Certificate interest = Total interest 0.07b + 0.02 170,000 - b2 = 2800 (Note that the units are consistent: the unit is dollars on each side.) STEP 4: 0.07b + 1400 - 0.02b = 2800 0.05b = 1400 b = 28,000 Candy should place $28,000 in the bonds and +70,000 - +28,000 = +42,000 in the certificate. STEP 5: The interest on the bonds after 1 year is 0.07 1+28,0002 = +1960; the interest on the certificate after 1 year is 0.021+42,0002 = +840. The total annual interest is $2800, the required amount. r Now Work PROBLEM 19 3 Solve Mixture Problems Oil refineries sometimes produce gasoline that is a blend of two or more types of fuel; bakeries occasionally blend two or more types of flour for their bread. These problems are referred to as mixture problems because they combine two or more quantities to form a mixture. EX AMPL E 4 Blending Coffees The manager of a Starbucks store decides to experiment with a new blend of coffee. She will mix some B grade Colombian coffee that sells for $5 per pound with some A grade Arabica coffee that sells for $10 per pound to get 100 pounds of the new blend. The selling price of the new blend is to be $7 per pound, and there is to be no difference in revenue between selling the new blend and selling the other types. How many pounds of the B grade Colombian coffee and how many pounds of the A grade Arabica coffees are required? Solution Let c represent the number of pounds of the B grade Colombian coffee. Then 100 - c equals the number of pounds of the A grade Arabica coffee. See Figure 16. $5 per pound $10 per pound + Figure 16 B Grade Colombian c pounds + $7 per pound = A Grade Arabica 100 − c pounds Blend = 100 pounds Since there is to be no difference in revenue between selling the A and B grades separately and selling the blend, we have Revenue from B grade + Revenue from A grade = Revenue from blend Price per pound Pounds of Price per pound Pounds of Price per pound Pounds of e fe f + e fe f = e fe f B grade A grade blend of B grade of A grade of blend # # 1100 - c2 = # c + +10 +7 100 +5 138 CHAPTER 1 Equations and Inequalities Now solve the equation: 5c + 101100 - c2 5c + 1000 - 10c - 5c c = = = = 700 700 - 300 60 The manager should blend 60 pounds of B grade Colombian coffee with 100 - 60 = 40 pounds of A grade Arabica coffee to get the desired blend. Check: The 60 pounds of B grade coffee would sell for 1+52 1602 = +300, and the 40 pounds of A grade coffee would sell for 1 +102 1402 = +400; the total revenue, $700, equals the revenue obtained from selling the blend, as desired. r Now Work PROBLEM 23 4 Solve Uniform Motion Problems Objects that move at a constant speed are said to be in uniform motion. When the average speed of an object is known, it can be interpreted as that object’s constant speed. For example, a bicyclist traveling at an average speed of 25 miles per hour can be considered in uniform motion with a constant speed of 25 miles per hour. Uniform Motion Formula If an object moves at an average speed (rate) r, the distance d covered in time t is given by the formula d = rt (2) That is, Distance = Rate # Time. EX AMPLE 5 Physics: Uniform Motion Tanya, who is a long-distance runner, runs at an average speed of 8 miles per hour (mi>h). Two hours after Tanya leaves your house, you leave in your Honda and follow the same route. If your average speed is 40 mi>h, how long will it be before you catch up to Tanya? How far will each of you be from your home? Solution Refer to Figure 17. We use t to represent the time (in hours) that it takes the Honda to catch up to Tanya. When this occurs, the total time elapsed for Tanya is t + 2 hours because she left 2 hours earlier. t50 2h Time t t50 Figure 17 Time t Set up the following table: Rate mi/h Time h Distance mi Tanya 8 t + 2 8(t + 2) Honda 40 t 40t SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 139 The distance traveled is the same for both, which leads to the equation 81t + 22 = 40t 8t + 16 = 40t 32t = 16 t = 1 hour 2 1 hour to catch up to Tanya. Each will have gone 20 miles. 2 1 Check: In 2.5 hours, Tanya travels a distance of 12.52 182 = 20 miles. In hour, 2 1 the Honda travels a distance of a b 1402 = 20 miles. 2 It will take the Honda r EXAMPL E 6 Physics: Uniform Motion A motorboat heads upstream a distance of 24 miles on a river whose current is running at 3 miles per hour (mi/h). The trip up and back takes 6 hours. Assuming that the motorboat maintained a constant speed relative to the water, what was its speed? Solution See Figure 18. Use r to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is r - 3 mi/h, and the true speed going downstream is r + 3 mi/h. Since Distance = Rate # Time, then Distance Time = . Set up a table. Rate 24 miles Rate mi/h Distance mi Distance Time = Rate h Upstream r - 3 24 24 r - 3 Downstream r + 3 24 24 r + 3 r 2 3 mi/h r 1 3 mi/h Figure 18 The total time up and back is 6 hours, which gives the equation 24 24 + = 6 r - 3 r + 3 241r + 32 + 241r - 32 = 6 1r - 32 1r + 32 48r = 6 r - 9 2 48r = 61r 2 - 92 6r 2 - 48r - 54 = 0 r - 8r - 9 = 0 2 1r - 92 1r + 12 = 0 r = 9 or r = - 1 Add the quotients on the left. Simplify. Multiply both sides by r2 - 9. Place in standard form. Divide by 6. Factor. Apply the Zero-Product Property and solve. Discard the solution r = - 1 mi/h and conclude that the speed of the motorboat relative to the water is 9 mi/h. r Now Work PROBLEM 29 140 CHAPTER 1 Equations and Inequalities 5 Solve Constant Rate Job Problems Here we look at jobs that are performed at a constant rate. The assumption is that 1 if a job can be done in t units of time, then of the job is done in 1 unit of time. In 1 t other words, if a job takes 4 hours, then of the job is done in 1 hour. 4 EX AMPLE 7 Solution Working Together to Do a Job At 10 am Danny is asked by his father to weed the garden. From past experience, Danny knows that this will take him 4 hours, working alone. His older brother Mike, when it is his turn to do this job, requires 6 hours. Since Mike wants to go golfing with Danny and has a reservation for 1 pm, he agrees to help Danny. Assuming no gain or loss of efficiency, when will they finish if they work together? Can they make the golf date? 1 1 Set up Table 2. In 1 hour, Danny does of the job, and in 1 hour, Mike does of the 4 6 job. Let t be the time (in hours) that it takes them to do the job together. In 1 hour, 1 then, of the job is completed. Reason as follows: t a Table 2 Hours to Do Job Part of Job Done in 1 Hour Danny 4 1 4 Mike 6 1 6 Together t 1 t Part done by Danny Part done by Mike Part done together b + a b = a b in 1 hour in 1 hour in 1 hour From Table 2, 1 1 + 4 6 3 2 + 12 12 5 12 5t 1 t 1 = t 1 = t = 12 = t = 12 5 The model. LCD = 12 on the left. Simplify. Multiply both sides by 12t. Divide each side by 5. 12 hours, or 2 hours, 24 minutes. 5 They should make the golf date, since they will finish at 12:24 pm. Working together, Mike and Danny can do the job in r Now Work PROBLEM 35 1.7 Assess Your Understanding Concepts and Vocabulary 1. The process of using variables to represent unknown quantities and then finding relationships that involve these variables is referred to as . 2. The money paid for the use of money is . 3. Objects that move at a constant speed are said to be in . 4. True or False The amount charged for the use of principal for a given period of time is called the rate of interest. 5. True or False If an object moves at an average speed r, the distance d covered in time t is given by the formula d = rt. 6. Suppose that you want to mix two coffees in order to obtain 100 pounds of a blend. If x represents the number of pounds of coffee A, write an algebraic expression that represents the number of pounds of coffee B. (a) 100 - x (b) x - 100 (c) 100 x (d) 100 + x 7. Which of the following is the simple interest formula? P rt (b) I = Prt (c) I = (d) I = P + rt (a) I = P rt 8. If it takes 5 hours to complete a job, what fraction of the job is done in 1 hour? 5 1 1 4 (a) (b) (c) (d) 5 4 5 4 SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications 141 Applications and Extensions In Problems 9–18, translate each sentence into a mathematical equation. Be sure to identify the meaning of all symbols. 9. Geometry The area of a circle is the product of the number p and the square of the radius. 10. Geometry The circumference of a circle is the product of the number p and twice the radius. 11. Geometry The area of a square is the square of the length of a side. 12. Geometry The perimeter of a square is four times the length of a side. 13. Physics Force equals the product of mass and acceleration. 14. Physics Pressure is force per unit area. 15. Physics Work equals force times distance. 16. Physics Kinetic energy is one-half the product of the mass and the square of the velocity. 17. Business The total variable cost of manufacturing x dishwashers is $150 per dishwasher times the number of dishwashers manufactured. 18. Business The total revenue derived from selling x dishwashers is $250 per dishwasher times the number of dishwashers sold. 25. Business: Mixing Nuts A nut store normally sells cashews for $9.00 per pound and almonds for $3.50 per pound. But at the end of the month the almonds had not sold well, so, in order to sell 60 pounds of almonds, the manager decided to mix the 60 pounds of almonds with some cashews and sell the mixture for $7.50 per pound. How many pounds of cashews should be mixed with the almonds to ensure no change in the revenue? 26. Business: Mixing Candy A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25 to produce and the cremes cost $0.45 to produce, how many of each should be in a box to yield a profit of $3? 27. Physics: Uniform Motion A motorboat can maintain a constant speed of 16 miles per hour relative to the water. The boat travels upstream to a certain point in 20 minutes; the return trip takes 15 minutes. What is the speed of the current? See the figure. 19. Financial Planning Betsy, a recent retiree, requires $6000 per year in extra income. She has $50,000 to invest and can invest in B-rated bonds paying 15% per year or in a certificate of deposit (CD) paying 7% per year. How much money should Betsy invest in each to realize exactly $6000 in interest per year? 20. Financial Planning After 2 years, Betsy (see Problem 19) finds that she will now require $7000 per year. Assuming that the remaining information is the same, how should the money be reinvested? 21. Banking A bank loaned out $12,000, part of it at the rate of 8% per year and the rest at the rate of 18% per year. If the interest received totaled $1000, how much was loaned at 8%? 22. Banking Wendy, a loan officer at a bank, has $1,000,000 to lend and is required to obtain an average return of 18% per year. If she can lend at the rate of 19% or at the rate of 16%, how much can she lend at the 16% rate and still meet her requirement? 23. Blending Teas The manager of a store that specializes in selling tea decides to experiment with a new blend. She will mix some Earl Grey tea that sells for $5 per pound with some Orange Pekoe tea that sells for $3 per pound to get 100 pounds of the new blend. The selling price of the new blend is to be $4.50 per pound, and there is to be no difference in revenue between selling the new blend and selling the other types. How many pounds of the Earl Grey tea and of the Orange Pekoe tea are required? 24. Business: Blending Coffee A coffee manufacturer wants to market a new blend of coffee that sells for $3.90 per pound by mixing two coffees that sell for $2.75 and $5 per pound, respectively. What amounts of each coffee should be blended to obtain the desired mixture? [Hint: Assume that the total weight of the desired blend is 100 pounds.] 28. Physics: Uniform Motion A motorboat heads upstream on a river that has a current of 3 miles per hour. The trip upstream takes 5 hours, and the return trip takes 2.5 hours. What is the speed of the motorboat? (Assume that the boat maintains a constant speed relative to the water.) 29. Physics: Uniform Motion A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. The total time for the trip was 1.5 hours. Use this information to find the speed of the current. 30. Physics: Uniform Motion Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car’s average speed is 10 miles per hour more than the other’s. The faster car arrives at 1 Wildwood at 11:00 am, hour before the other car. What 2 was the average speed of each car? How far did each travel? 31. Moving Walkways The speed of a moving walkway is typically about 2.5 feet per second. Walking on such a moving walkway, it takes Karen a total of 40 seconds to travel 50 feet with the movement of the walkway and then back again against the movement of the walkway. What is Karen’s normal walking speed? Source: Answers.com 36. Working Together on a Job Patrice, by himself, can paint four rooms in 10 hours. If he hires April to help, they can do the same job together in 6 hours. If he lets April work alone, how long will it take her to paint four rooms? 37. Enclosing a Garden A gardener has 46 feet of fencing to be used to enclose a rectangular garden that has a border 2 feet wide surrounding it. See the figure. (a) If the length of the garden is to be twice its width, what will be the dimensions of the garden? (b) What is the area of the garden? (c) If the length and width of the garden are to be the same, what will be the dimensions of the garden? (d) What will be the area of the square garden? 2 ft 2 ft 38. Construction A pond is enclosed by a wooden deck that is 3 feet wide. The fence surrounding the deck is 100 feet long. (a) If the pond is square, what are its dimensions? (b) If the pond is rectangular and the length of the pond is to be three times its width, what are its dimensions? (c) If the pond is circular, what is its diameter? (d) Which pond has the larger area? 39. Football A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20-yard line with the defensive back at the 15-yard line. (See the figure-top, right.) If no other players are nearby, at what yard line will the defensive back catch up to the tight end? [Hint: At time t = 0, the defensive back is 5 yards behind the tight end.] 30 20 TE DB 10 10 35. Working Together on a Job Trent can deliver his newspapers in 30 minutes. It takes Lois 20 minutes to do the same route. How long would it take them to deliver the newspapers if they worked together? 20 34. Laser Printers It takes an HP LaserJet M451dw laser printer 16 minutes longer to complete an 840-page print job by itself than it takes an HP LaserJet CP4025dn to complete the same job by itself. Together the two printers can complete the job in 15 minutes. How long does it take each printer to complete the print job alone? What is the speed of each printer? Source: Hewlett-Packard 30 33. Tennis A regulation doubles tennis court has an area of 2808 square feet. If it is 6 feet longer than twice its width, determine the dimensions of the court. Source: United States Tennis Association 40 32. High-Speed Walkways Toronto’s Pearson International Airport has a high-speed version of a moving walkway. If Liam walks while riding this moving walkway, he can travel 280 meters in 60 seconds less time than if he stands still on the moving walkway. If Liam walks at a normal rate of 1.5 meters per second, what is the speed of the walkway? Source: Answers.com 40 CHAPTER 1 Equations and Inequalities SOUTH 142 40. Computing Business Expense Therese, an outside salesperson, uses her car for both business and pleasure. Last year, she traveled 30,000 miles, using 900 gallons of gasoline. Her car gets 40 miles per gallon on the highway and 25 in the city. She can deduct all highway travel, but no city travel, on her taxes. How many miles should Therese deduct as a business expense? 41. Mixing Water and Antifreeze How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze? 42. Mixing Water and Antifreeze The cooling system of a certain foreign-made car has a capacity of 15 liters. If the system is filled with a mixture that is 40% antifreeze, how much of this mixture should be drained and replaced by pure antifreeze so that the system is filled with a solution that is 60% antifreeze? 43. Chemistry: Salt Solutions How much water must be evaporated from 32 ounces of a 4% salt solution to make a 6% salt solution? 44. Chemistry: Salt Solutions How much water must be evaporated from 240 gallons of a 3% salt solution to produce a 5% salt solution? 45. Purity of Gold The purity of gold is measured in karats, with pure gold being 24 karats. Other purities of gold are expressed as proportional parts of pure gold. Thus, 18-karat 18 12 gold is , or 75% pure gold; 12-karat gold is , or 50% 24 24 pure gold; and so on. How much 12-karat gold should be mixed with pure gold to obtain 60 grams of 16-karat gold? 46. Chemistry: Sugar Molecules A sugar molecule has twice as many atoms of hydrogen as it does oxygen and one more atom of carbon than of oxygen. If a sugar molecule has a total of 45 atoms, how many are oxygen? How many are hydrogen? 47. Running a Race Mike can run the mile in 6 minutes, and Dan can run the mile in 9 minutes. If Mike gives Dan a head start of 1 minute, how far from the start will Mike pass Dan? How long does it take? See the figure. Dan Mike Start 1– 4 1– 2 mi mi 3– 4 mi Chapter Review 48. Range of an Airplane An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 mi/h, how far can it fly and return safely? (Assume that the wind remains constant.) 49. Emptying Oil Tankers An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 9 hours. If the main pump is started at 9 am, when should the auxiliary pump be started so that the tanker is emptied by noon? 50. Cement Mix A 20-pound bag of Economy brand cement mix contains 25% cement and 75% sand. How much pure cement must be added to produce a cement mix that is 40% cement? 51. Filling a Tub A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both faucets closed and the stopper removed, the tub will empty in 20 minutes. How long will it take for the tub to fill if both faucets are open and the stopper is removed? 52. Using Two Pumps A 5-horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool. After two hours, the 2-hp pump breaks down. How long will it take the larger pump to finish emptying the pool? 53. A Biathlon Suppose that you have entered an 87-mile biathlon that consists of a run and a bicycle race. During 143 your run, your average speed is 6 miles per hour, and during your bicycle race, your average speed is 25 miles per hour. You finish the race in 5 hours. What is the distance of the run? What is the distance of the bicycle race? 54. Cyclists Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist. After 6 hours they are 246 miles apart. How fast is each cyclist riding? 55. Comparing Olympic Heroes In the 2012 Olympics, Usain Bolt of Jamaica won the gold medal in the 100-meter race with a time of 9.69 seconds. In the 1896 Olympics, Thomas Burke of the United States won the gold medal in the 100-meter race in 12.0 seconds. If they ran in the same race, repeating their respective times, by how many meters would Bolt beat Burke? 56. Constructing a Coffee Can A 39-ounce can of Hills Bros.® coffee requires 188.5 square inches of aluminum. If its height is 7 inches, what is its radius? [Hint: The surface area S of a right cylinder is S = 2pr 2 + 2prh, where r is the radius and h is the height.] 7 in. 39 oz. Explaining Concepts: Discussion and Writing 57. Critical Thinking You are the manager of a clothing store and have just purchased 100 dress shirts for $20.00 each. After 1 month of selling the shirts at the regular price, you plan to have a sale giving 40% off the original selling price. However, you still want to make a profit of $4 on each shirt at the sale price. What should you price the shirts at initially to ensure this? If, instead of 40% off at the sale, you give 50% off, by how much is your profit reduced? 60. Computing Average Speed In going from Chicago to Atlanta, a car averages 45 miles per hour, and in going from Atlanta to Miami, it averages 55 miles per hour. If Atlanta is halfway between Chicago and Miami, what is the average speed from Chicago to Miami? Discuss an intuitive solution. Write a paragraph defending your intuitive solution. Then solve the problem algebraically. Is your intuitive solution the same as the algebraic one? If not, find the flaw. 58. Critical Thinking Make up a word problem that requires solving a linear equation as part of its solution. Exchange problems with a friend. Write a critique of your friend’s problem. 61. Speed of a Plane On a recent flight from Phoenix to Kansas City, a distance of 919 nautical miles, the plane arrived 20 minutes early. On leaving the aircraft, I asked the captain, “What was our tail wind?” He replied, “I don’t know, but our ground speed was 550 knots.” Has enough information been provided for you to find the tail wind? If possible, find the tail wind. (1 knot = 1 nautical mile per hour) 59. Critical Thinking Without solving, explain what is wrong with the following mixture problem: How many liters of 25% ethanol should be added to 20 liters of 48% ethanol to obtain a solution of 58% ethanol? Now go through an algebraic solution. What happens? Chapter Review Things to Know Quadratic formula (pp. 97 and 110) - b { 2b2 - 4ac . 2a If b2 - 4ac 6 0, there are no real solutions. If ax2 + bx + c = 0, a ≠ 0, then x = Discriminant (pp. 97 and 110) If b2 - 4ac 7 0, there are two unequal real solutions. If b2 - 4ac = 0, there is one repeated real solution, a double root. If b2 - 4ac 6 0, there are no real solutions, but there are two distinct complex solutions that are not real; the complex solutions are conjugates of each other. 144 CHAPTER 1 Equations and Inequalities Interval notation (p. 120) 3a, b4 5x a … x … b6 3a, q 2 5x x Ú a6 1a, b4 5x a 6 x … b6 1 - q , a4 5x x … a6 3a, b2 5x a … x 6 b6 1a, b2 1a, q 2 1 - q , a2 5x a 6 x 6 b6 1 - q, q 2 5x x 7 a6 5x x 6 a6 All real numbers Properties of inequalities Addition property (p. 121) If a 6 b, then a + c 6 b + c. Multiplication properties (p. 122) (a) If a 6 b and if c 7 0, then ac 6 bc. (b) If a 7 b and if c 7 0, then ac 7 bc. If a 6 b and if c 6 0, then ac 7 bc. If a 7 b and if c 6 0, then ac 6 bc. If a 7 b, then a + c 7 b + c. Reciprocal properties (p. 123) If a 7 0, then If 1 7 0. a 1 7 0, then a 7 0. a If a 6 0, then If 1 6 0. a 1 6 0, then a 6 0. a Absolute value If 0 u 0 = a, a 7 0, then u = - a or u = a. (p. 130) If 0 u 0 … a, a 7 0, then - a … u … a. (p. 131) If 0 u 0 Ú a, a 7 0, then u … - a or u Ú a. (p. 132) Objectives Section 1.1 1 2 3 1.2 1 2 3 4 1.3 1 2 1.4 1 2 3 1.5 1 2 3 4 1.6 1 2 1.7 1 2 3 4 5 You should be able to . . . Examples Review Exercises Solve a linear equation (p. 84) Solve equations that lead to linear equations (p. 86) Solve problems that can be modeled by linear equations (p. 87) 1–3 4–6 8, 9 1–3, 7, 8 4 45, 57 Solve a quadratic equation by factoring (p. 93) Solve a quadratic equation by completing the square (p. 95) Solve a quadratic equation using the quadratic formula (p. 96) Solve problems that can be modeled by quadratic equations (p. 99) 1, 2 4, 5 6–9 10 6, 9, 21, 22 5, 6, 9, 10, 13, 21, 22 5, 6, 9, 10, 13, 21, 22 50, 54, 56 Add, subtract, multiply, and divide complex numbers (p. 105) Solve quadratic equations in the complex number system (p. 109) 1–8 9–12 35–39 40–43 Solve radical equations (p. 113) Solve equations quadratic in form (p. 114) Solve equations by factoring (p. 116) 1–3 4–6 7, 8 11, 12, 15–19, 23 14, 20 26, 27 Use interval notation (p. 120) Use properties of inequalities (p. 121) Solve inequalities (p. 123) Solve combined inequalities (p. 124) 1, 2 3–6 7, 8 9, 10 28–34 28–34, 47 28, 47 29, 30 Solve equations involving absolute value (p. 130) Solve inequalities involving absolute value (p. 130) 1 2–6 24, 25 31–34 Translate verbal descriptions into mathematical expressions (p. 135) Solve interest problems (p. 136) Solve mixture problems (p. 137) Solve uniform motion problems (p. 138) Solve constant rate job problems (p. 140) 1 2, 3 4 5, 6 7 44 45 53, 55 46, 48, 49, 59, 60 51, 52, 58 Chapter Review 145 Review Exercises In Problems 1–27, find the real solutions, if any, of each equation. (Where they appear, a, b, m, and n are positive constants.) 1. 2 - x = 8 3 4. x 6 = x - 1 5 7. 1 1 3 x ax - b = 2 3 4 6 x ≠ 1 x 1 3x - = 4 3 12 2. - 215 - 3x2 + 8 = 4 + 5x 3. 5. x11 - x2 = 6 6. x11 + x2 = 6 8. 9. 1x - 12 12x + 32 = 3 1 - 3x x + 6 1 = + 4 3 2 10. 2x + 3 = 4x2 3 2 11. 2 x - 1 = 2 12. 21 + x3 = 3 13. x1x + 12 + 2 = 0 14. x4 - 5x2 + 4 = 0 15. 22x - 3 + x = 3 4 16. 2 2x + 3 = 2 17. 2x + 1 + 2x - 1 = 22x + 1 18. 22x - 1 - 2x - 5 = 3 19. 2x1>2 - 3 = 0 20. x -6 - 7x -3 - 8 = 0 21. x2 + m2 = 2mx + 1nx2 2 n ≠ 1 23. 2x2 + 3x + 7 - 2x2 - 3x + 9 + 2 = 0 22. 10a2 x2 - 2abx - 36b2 = 0 24. 0 2x + 3 0 = 7 25. 0 2 - 3x 0 + 2 = 9 26. 2x3 = 3x2 27. 2x3 + 5x2 - 8x - 20 = 0 In Problems 28–34, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. 28. 2x - 3 x + 2 … 5 2 32. 0 2x - 5 0 Ú 9 29. - 9 … 2x + 3 … 7 -4 33. 2 + 0 2 - 3x 0 … 4 30. 2 6 31. 0 3x + 4 0 6 3 - 3x 6 6 12 1 2 34. 1 - 0 2 - 3x 0 6 - 4 In Problems 35–39, use the complex number system and write each expression in the standard form a + bi. 35. 16 + 3i2 - 12 - 4i2 36. 413 - i2 + 31 - 5 + 2i2 38. i 50 39. 12 + 3i2 3 37. 3 3 + i In Problems 40–43, solve each equation in the complex number system. 40. x2 + x + 1 = 0 41. 2x2 + x - 2 = 0 44. Translate the following statement into a mathematical expression: The total cost C of manufacturing x bicycles in one day is $50,000 plus $95 times the number of bicycles manufactured. 45. Financial Planning Steve, a recent retiree, requires $5000 per year in extra income. He has $70,000 to invest and can invest in A-rated bonds paying 8% per year or in a certificate of deposit (CD) paying 5% per year. How much money should be invested in each to realize exactly $5000 in interest per year? 46. Lightning and Thunder A flash of lightning is seen, and the resulting thunderclap is heard 3 seconds later. If the speed of sound averages 1100 feet per second, how far away is the storm? 42. x2 + 3 = x 43. x11 - x2 = 6 146 CHAPTER 1 Equations and Inequalities 47. Physics: Intensity of Light The intensity I (in candlepower) 900 of a certain light source obeys the equation I = 2 , where x x is the distance (in meters) from the light. Over what range of distances can an object be placed from this light source so that the range of intensity of light is from 1600 to 3600 candlepower, inclusive? 53. Chemistry: Salt Solutions How much water should be added to 64 ounces of a 10% salt solution to make a 2% salt solution? 54. Geometry The diagonal of a rectangle measures 10 inches. If the length is 2 inches more than the width, find the dimensions of the rectangle. 55. Chemistry: Mixing Acids A laboratory has 60 cubic centimeters (cm3) of a solution that is 40% HCl acid. How many cubic centimeters of a 15% solution of HCl acid should be mixed with the 60 cm3 of 40% acid to obtain a solution of 25% HCl? How much of the 25% solution is there? 48. Extent of Search and Rescue A search plane has a cruising speed of 250 miles per hour and carries enough fuel for at most 5 hours of flying. If there is a wind that averages 30 miles per hour and the direction of the search is with the wind one way and against it the other, how far can the search plane travel before it has to turn back? 56. Framing a Painting An artist has 50 inches of oak trim to frame a painting. The frame is to have a border 3 inches wide surrounding the painting. (a) If the painting is square, what are its dimensions? What are the dimensions of the frame? (b) If the painting is rectangular with a length twice its width, what are the dimensions of the painting? What are the dimensions of the frame? 49. Rescue at Sea A life raft, set adrift from a sinking ship 150 miles offshore, travels directly toward a Coast Guard station at the rate of 5 miles per hour. At the time that the raft is set adrift, a rescue helicopter is dispatched from the Coast Guard station. If the helicopter’s average speed is 90 miles per hour, how long will it take the helicopter to reach the life raft? 57. Finance An inheritance of $900,000 is to be divided among Scott, Alice, and Tricia in the following manner: Alice is to 1 3 receive of what Scott gets, while Tricia gets of what Scott 4 2 gets. How much does each receive? 90 mi/h 58. Utilizing Copying Machines A new copying machine can do a certain job in 1 hour less than an older copier. Together they can do this job in 72 minutes. How long would it take the older copier by itself to do the job? 5 mi/h 150 mi 59. Evening Up a Race In a 100-meter race, Todd crosses the finish line 5 meters ahead of Scott. To even things up, Todd suggests to Scott that they race again, this time with Todd lining up 5 meters behind the start. (a) Assuming that Todd and Scott run at the same pace as before, does the second race end in a tie? (b) If not, who wins? 50. Physics An object is thrown down from the top of a building 1280 feet tall with an initial velocity of 32 feet per second. The distance s (in feet) of the object from the ground after t seconds is s = 1280 - 32t - 16t 2. (a) When will the object strike the ground? (b) What is the height of the object after 4 seconds? (c) By how many meters does he win? (d) How far back should Todd start so that the race ends in a tie? 51. Working Together to Get a Job Done Clarissa and Shawna, working together, can paint the exterior of a house in 6 days. Clarissa by herself can complete this job in 5 days less than Shawna. How long will it take Clarissa to complete the job by herself? After running the race a second time, Scott, to even things up, suggests to Todd that he (Scott) line up 5 meters in front of the start. (e) Assuming again that they run at the same pace as in the first race, does the third race result in a tie? (f) If not, who wins? (g) By how many meters? (h) How far ahead should Scott start so that the race ends in a tie? 52. Emptying a Tank Two pumps of different sizes, working together, can empty a fuel tank in 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger pump is out of order, how long will it take the smaller one to do the job alone? 60. Physics: Uniform Motion A man is walking at an average speed of 4 miles per hour alongside a railroad track. A freight train, going in the same direction at an average speed of 30 miles per hour, requires 5 seconds to pass the man. How long is the freight train? Give your answer in feet. 30 mi/h Length of train 4 mi/h t50 5 sec t55 Chapter Projects The Chapter Test Prep Videos are step-by-step solutions available in , or on this text’s Channel. Flip back to the Resources for Success page for a link to this text’s YouTube channel. Chapter Test In Problems 1–7, solve each equation. 2x x 5 2. x1x - 12 = 6 1. - = 3 2 12 5. 0 2x - 3 0 + 7 = 10 147 6. 3x3 + 2x2 - 12x - 8 = 0 4. 22x - 5 + 2 = 4 3. x4 - 3x2 - 4 = 0 7. 3x2 - x + 1 = 0 In Problems 8–10, solve each inequality. Express your answer using interval notation. Graph the solution set. 8. - 3 … 11. Write 3x - 4 … 6 2 9. 0 3x + 4 0 6 8 10. 2 + 0 2x - 5 0 Ú 9 -2 in the standard form a + bi. 3 - i 12. Solve the equation 4x2 - 4x + 5 = 0 in the complex number system. 13. Blending Coffee A coffee house has 20 pounds of a coffee that sells for $4 per pound. How many pounds of a coffee that sells for $8 per pound should be mixed with the 20 pounds of $4-per-pound coffee to obtain a blend that will sell for $5 per pound? How much of the $5-per-pound coffee is there to sell? Chapter Projects time t, measured in months, is the length of the loan. For example, a 30-year loan requires 12 * 30 = 360 monthly payments. P = L≥ Internet-based Project I. Financing a Purchase At some point in your life, you are likely to need to borrow money to finance a purchase. For example, most of us will finance the purchase of a car or a home. What is the mathematics behind financing a purchase? When you borrow money from a bank, the bank uses a rather complex equation (or formula) to determine how much you need to pay each month to repay the loan. There are a number of variables that determine the monthly payment. These variables include the amount borrowed, the interest rate, and the length of the loan. The interest rate is based on current economic conditions, the length of the loan, the type of item being purchased, and your credit history. The following formula gives the monthly payment P required to pay off a loan amount L at an annual interest rate r, expressed as a decimal, but usually given as a percent. The r 12 1 - a1 + r -t b 12 ¥ P = monthly payment L = loan amount r = annual rate of interest expressed as a decimal t = length of loan, in months 1. Interest rates change daily. Many websites post current interest rates on loans. Go to www.bankrate.com (or some other website that posts lenders’ interest rates) and find the current best interest rate on a 60-month new-car purchase loan. Use this rate to determine the monthly payment on a $30,000 automobile loan. 2. Determine the total amount paid for the loan by multiplying the loan payment by the term of the loan. 3. Determine the total amount of interest paid by subtracting the loan amount from the total amount paid from question 2. 4. More often than not, we decide how much of a payment we can afford and use that information to determine the loan amount. Suppose you can afford a monthly payment of $500. Use the interest rate from question 1 to determine the maximum amount you can borrow. If you have $5000 to put down on the car, what is the maximum value of a car you can purchase? 5. Repeat questions 1 through 4 using a 72-month new-car purchase loan, a 60-month used-car purchase loan, and a 72-month used-car purchase loan. 148 CHAPTER 1 Equations and Inequalities 6. We can use the power of a spreadsheet, such as Excel, to create a loan amortization schedule. A loan amortization schedule is a list of the monthly payments, a breakdown of interest and principal, along with a current loan balance. Create a loan amortization schedule for each of the four loan scenarios discussed on the previous page, using the following as a guide. You may want to use an Internet search engine to research specific keystrokes for creating an amortization schedule in a spreadsheet. We supply a sample spreadsheet with formulas included as a guide. Use the spreadsheet to verify your results from questions 1 through 5. Loan Information Loan Amount Annual Interest Rate Length of Loan (years) Number of Payments 7. $30,000.00 0.045 5 =B4*12 Payment Number Payment Amount Interest Principal Balance Total Interest Paid 1 2 3 · · · =PMT($B$3/12,$B$5,−$B$2,0) =PMT($B$3/12,$B$5,−$B$2,0) =PMT($B$3/12,$B$5,−$B$2,0) · · · =B2*$B$3/12 =H2*$B$3/12 =H3*$B$3/12 · · · =E2−F2 =E3−F3 =E4−F4 · · · =B2−G2 =H2−G3 =H3−G4 · · · =B2*$B$3/12 =I2+F3 =I3+F4 · · · Go to an online automobile website such as www.cars.com, www.edmunds.com, or www.autobytel.com. Research the types of vehicles you can afford for a monthly payment of $500. Decide on a vehicle you would purchase based on your analysis in questions 1–6. Be sure to justify your decision, and include the impact the term of the loan has on your decision. You might consider other factors in your decision, such as expected maintenance costs and insurance costs. Citation: Excel ©2013 Microsoft Corporation. Used with permission from Microsoft. The following project is also available on the Instructor’s Resource Center (IRC): II. Project at Motorola How Many Cellular Phones Can I Make? An industrial engineer uses a model involving equations to be sure production levels meet customer demand. 2 Graphs How to Value a House Two things to consider in valuing a home are, first, how does it compare to similar homes that have sold recently? Is the asking price fair? And second, what value do you place on the advertised features and amenities? Yes, other people might value them highly, but do you? Zestimate home valuation, RealEstateABC.com, and Reply.com are among the many algorithmic (generated by a computer model) starting points in figuring out the value of a home. They show you how the home is priced relative to other homes in the area, but you need to add in all the things that only someone who has seen the house knows. You can do that using My Estimator, and then you create your own estimate and see how it stacks up against the asking price. Looking at “Comps” Knowing whether an asking price is fair will be important when you’re ready to make an offer on a house. It will be even more important when your mortgage lender hires an appraiser to determine whether the house is worth the loan you want. Check with your agent, Zillow.com, propertyshark.com, or other websites to see recent sales of homes in the area that are similar, or comparable, to what you’re looking for. Print them out and keep these “comps” in a three-ring binder; you’ll be referring to them quite a bit. Note that “recent sales” usually means within the last six months. A sales price from a year ago may bear little or no relation to what is going on in your area right now. In fact, some lenders will not accept comps older than three months. Market activity also determines how easy or difficult it is to find accurate comps. In a “hot” or busy market, with sales happening all the time, you’re likely to have lots of comps to choose from. In a less active market, finding reasonable comps becomes harder. And if the home you’re looking at has special design features, finding a comparable property is harder still. It’s also necessary to know what’s going on in a given sub-segment. Maybe large, high-end homes are selling like hotcakes, but owners of smaller houses are staying put, or vice versa. Source: http://allmyhome.blogspot.com/2008/07/how-to-value-house.html —See the Internet-based Chapter Project— A Look Back Chapter R and Chapter 1 review skills from intermediate algebra. A Look Ahead Here we connect algebra and geometry using the rectangular coordinate system. In the 1600s, algebra had developed to the point that René Descartes (1596–1650) and Pierre de Fermat (1601–1665) were able to use rectangular coordinates to translate geometry problems into algebra problems, and vice versa. This enabled both geometers and algebraists to gain new insights into their subjects, which had been thought to be separate but now were seen as connected. Outline 2.1 2.2 2.3 2.4 2.5 The Distance and Midpoint Formulas Graphs of Equations in Two Variables; Intercepts; Symmetry Lines Circles Variation Chapter Review Chapter Test Cumulative Review Chapter Project 149 149 150 CHAPTER 2 Graphs 2.1 The Distance and Midpoint Formulas PREPARING FOR THIS SECTION Before getting started, review the following: r Algebra Essentials (Chapter R, Section R.2, pp. 17–26) r Geometry Essentials (Chapter R, Section R.3, pp. 30–35) Now Work the ‘Are You Prepared?’ problems on page 154. OBJECTIVES 1 Use the Distance Formula (p. 151) 2 Use the Midpoint Formula (p. 153) Rectangular Coordinates y 4 2 –4 –2 O 2 4 x –2 –4 Figure 1 xy@Plane y 4 3 (–3, 1) 1 –4 3 (–2, –3) 3 (3, 2) 2 O 3 x 4 2 (3, –2) 2 Figure 2 y Quadrant II x < 0, y > 0 Quadrant I x > 0, y > 0 Quadrant III x < 0, y < 0 Quadrant IV x > 0, y < 0 x Figure 3 We locate a point on the real number line by assigning it a single real number, called the coordinate of the point. For work in a two-dimensional plane, we locate points by using two numbers. Begin with two real number lines located in the same plane: one horizontal and the other vertical. The horizontal line is called the x-axis, the vertical line the y-axis, and the point of intersection the origin O. See Figure 1. Assign coordinates to every point on these number lines using a convenient scale. In mathematics, we usually use the same scale on each axis, but in applications, different scales appropriate to the application may be used. The origin O has a value of 0 on both the x-axis and the y-axis. Points on the x-axis to the right of O are associated with positive real numbers, and those to the left of O are associated with negative real numbers. Points on the y-axis above O are associated with positive real numbers, and those below O are associated with negative real numbers. In Figure 1, the x-axis and y-axis are labeled as x and y, respectively, and an arrow at the end of each axis is used to denote the positive direction. The coordinate system described here is called a rectangular or Cartesian* coordinate system. The plane formed by the x-axis and y-axis is sometimes called the xy-plane, and the x-axis and y-axis are referred to as the coordinate axes. Any point P in the xy-plane can be located by using an ordered pair 1x, y2 of real numbers. Let x denote the signed distance of P from the y-axis (signed means that if P is to the right of the y-axis, then x 7 0, and if P is to the left of the y-axis, then x 6 0); and let y denote the signed distance of P from the x-axis. The ordered pair 1x, y2 , also called the coordinates of P, gives us enough information to locate the point P in the plane. For example, to locate the point whose coordinates are 1 - 3, 12, go 3 units along the x-axis to the left of O and then go straight up 1 unit. We plot this point by placing a dot at this location. See Figure 2, in which the points with coordinates 1 - 3, 12 , 1 - 2, - 32, 13, - 22 , and 13, 22 are plotted. The origin has coordinates 10, 02 . Any point on the x-axis has coordinates of the form 1x, 02, and any point on the y-axis has coordinates of the form 10, y2 . If 1x, y2 are the coordinates of a point P, then x is called the x-coordinate, or abscissa, of P, and y is the y-coordinate, or ordinate, of P. We identify the point P by its coordinates 1x, y2 by writing P = 1x, y2 . Usually, we will simply say “the point 1x, y2 ” rather than “the point whose coordinates are 1x, y2 .” The coordinate axes divide the xy-plane into four sections called quadrants, as shown in Figure 3. In quadrant I, both the x-coordinate and the y-coordinate of all points are positive; in quadrant II, x is negative and y is positive; in quadrant III, both x and y are negative; and in quadrant IV, x is positive and y is negative. Points on the coordinate axes belong to no quadrant. Now Work PROBLEM 15 * Named after René Descartes (1596–1650), a French mathematician, philosopher, and theologian. SECTION 2.1 The Distance and Midpoint Formulas 151 COMMENT On a graphing calculator, you can set the scale on each axis. Once this has been done, you obtain the viewing rectangle. See Figure 4 for a typical viewing rectangle. You should now read Section 1, The Viewing Rectangle, in the Appendix. Figure 4 TI-84 Plus C Standard Viewing Rectangle ■ 1 Use the Distance Formula If the same units of measurement (such as inches, centimeters, and so on) are used for both the x-axis and y-axis, then all distances in the xy-plane can be measured using this unit of measurement. EXAM PL E 1 Solution Finding the Distance between Two Points Find the distance d between the points 11, 32 and 15, 62 . First plot the points 11, 32 and 15, 62 and connect them with a straight line. See Figure 5(a). To find the length d, begin by drawing a horizontal line from 11, 32 to 15, 32 and a vertical line from 15, 32 to 15, 62, forming a right triangle, as shown in Figure 5(b). One leg of the triangle is of length 4 (since 0 5 - 1 0 = 4), and the other is of length 3 (since 0 6 - 3 0 = 3). By the Pythagorean Theorem, the square of the distance d that we seek is d 2 = 42 + 32 = 16 + 9 = 25 d = 225 = 5 y 6 y 6 (5, 6) d d 3 (5, 6) 3 (1, 3) 6 x 3 3 (1, 3) 4 (5, 3) 3 6 x (b) (a) Figure 5 r The distance formula provides a straightforward method for computing the distance between two points. THEOREM Distance Formula The distance between two points P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2, denoted by d 1P1 , P2 2, is In Words To compute the distance between two points, find the difference of the x-coordinates, square it, and add this to the square of the difference of the y-coordinates. The square root of this sum is the distance. d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2 (1) Proof of the Distance Formula Let 1x1 , y1 2 denote the coordinates of point P1 and let 1x2 , y2 2 denote the coordinates of point P2. Assume that the line joining P1 and P2 is neither horizontal nor vertical. Refer to Figure 6(a) on page 152. The coordinates of P3 are 1x2 , y1 2 . The horizontal distance from P1 to P3 is the absolute 152 CHAPTER 2 Graphs value of the difference of the x-coordinates, 0 x2 - x1 0 . The vertical distance from P3 to P2 is the absolute value of the difference of the y-coordinates, 0 y2 - y1 0 . See Figure 6(b). The distance d 1P1 , P2 2 is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that 3 d 1P1 , P2 2 4 2 = 0 x2 - x1 0 2 + 0 y2 - y1 0 2 = 1x2 - x1 2 2 + 1y2 - y1 2 2 d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2 y y y2 P2 (x2, y2) y1 P3 (x2, y1) P1 (x1, y1) x2 x1 P2 (x2, y2) y2 d(P1, P2) y1 P1 (x1, y1) x ⏐y2 y1⏐ ⏐x2 x1⏐ x1 x2 P3 (x2, y1) x (b) (a) Figure 6 Now, if the line joining P1 and P2 is horizontal, then the y-coordinate of P1 equals the y-coordinate of P2; that is, y1 = y2 . Refer to Figure 7(a). In this case, the distance formula (1) still works, because for y1 = y2 , it reduces to d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 02 = 2 1x2 - x1 2 2 = 0 x2 - x1 0 y y y2 P1 (x1, y1) y1 d (P1, P2) P2 (x2, y1) P1 (x1, y1) y1 ⏐x2 x1⏐ x1 P2 (x1, y2) ⏐y2 y1⏐ d (P1, P2) x2 x1 x (a) x (b) Figure 7 A similar argument holds if the line joining P1 and P2 is vertical. See Figure 7(b). ■ EX AMPLE 2 Solution Using the Distance Formula Find the distance d between the points 1 - 4, 52 and (3, 2). Using the distance formula, equation (1), reveals that the distance d is d = 2 3 3 - 1 - 42 4 2 + 12 - 52 2 = 272 + 1 - 32 2 = 249 + 9 = 258 ≈ 7.62 Now Work PROBLEMS 19 AND 23 r The distance between two points P1 = 1x1, y1 2 and P2 = 1x2, y2 2 is never a negative number. Also, the distance between two points is 0 only when the points are identical—that is, when x1 = x2 and y1 = y2. And, because 1x2 - x1 2 2 = 1x1 - x2 2 2 and 1y2 - y1 2 2 = 1y1 - y2 2 2, it makes no difference whether the distance is computed from P1 to P2 or from P2 to P1; that is, d 1P1 , P2 2 = d 1P2 , P1 2 . The introduction to this chapter mentioned that rectangular coordinates enable us to translate geometry problems into algebra problems, and vice versa. The next example shows how algebra (the distance formula) can be used to solve geometry problems. SECTION 2.1 The Distance and Midpoint Formulas 153 Using Algebra to Solve Geometry Problems EXAMPL E 3 Consider the three points A = 1 - 2, 12, B = 12, 32, and C = 13, 12 . (a) (b) (c) (d) Solution Plot each point and form the triangle ABC. Find the length of each side of the triangle. Show that the triangle is a right triangle. Find the area of the triangle. (a) Figure 8 shows the points A, B, C and the triangle ABC. (b) To find the length of each side of the triangle, use the distance formula, equation (1). d 1A, B2 = 2 3 2 - 1 - 22 4 2 + 13 - 12 2 = 216 + 4 = 220 = 225 y d 1B, C2 = 2 13 - 22 2 + 11 - 32 2 = 21 + 4 = 25 B = (2, 3) 3 A = (–2, 1) d 1A, C2 = 2 3 3 - 1 - 22 4 2 + 11 - 12 2 = 225 + 0 = 5 C = (3, 1) –3 (c) If the sum of the squares of the lengths of two of the sides equals the square of the length of the third side, then the triangle is a right triangle. Looking at Figure 8, it seems reasonable to conjecture that the angle at vertex B might be a right angle. We shall check to see whether x 3 Figure 8 3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 3 d 1A, C2 4 2 Using the results in part (b) yields 3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 1 225 2 2 + 1 25 2 2 = 20 + 5 = 25 = 3 d 1A, C2 4 2 It follows from the converse of the Pythagorean Theorem that triangle ABC is a right triangle. (d) Because the right angle is at vertex B, the sides AB and BC form the base and height of the triangle. Its area is Area = 1 1 1Base2 1Height2 = 1 225 2 1 25 2 = 5 square units 2 2 Now Work PROBLEM r 31 2 Use the Midpoint Formula y P2 = (x 2, y2) y2 M = (x, y) y y1 y – y1 x – x1 P1 = (x1, y1) x1 Figure 9 y2 – y x2 – x B = (x 2, y) x - x1 = x2 - x A = (x, y1) x We now derive a formula for the coordinates of the midpoint of a line segment. Let P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2 be the endpoints of a line segment, and let M = 1x, y2 be the point on the line segment that is the same distance from P1 as it is from P2 . See Figure 9. The triangles P1 AM and MBP2 are congruent. [Do you see why? d 1P1 , M2 = d 1M, P2 2 is given; also, ∠AP1 M = ∠BMP2* and ∠P1 MA = ∠MP2 B. Thus, we have angle–side–angle.] Because triangles P1 AM and MBP2 are congruent, corresponding sides are equal in length. That is, x2 x 2x = x1 + x2 x = x1 + x2 2 and y - y1 = y2 - y 2y = y1 + y2 y = y1 + y2 2 *A postulate from geometry states that the transversal P1P2 forms congruent corresponding angles with the parallel line segments P1A and MB. 154 CHAPTER 2 Graphs THEOREM Midpoint Formula The midpoint M = 1x, y2 of the line segment from P1 = 1x1 , y1 2 to P2 = 1x2, y2 2 is In Words M = 1x, y2 = ¢ To find the midpoint of a line segment, average the x-coordinates of the endpoints, and average the y-coordinates of the endpoints. Find the midpoint of the line segment from P1 = 1 - 5, 52 to P2 = 13, 12 . Plot the points P1 and P2 and their midpoint. Solution y M (–1, 3) –5 Apply the midpoint formula (2) using x1 = - 5, y1 = 5, x2 = 3, and y2 = 1. Then the coordinates 1x, y2 of the midpoint M are x = 5 y1 + y2 x1 + x2 -5 + 3 5 + 1 = = - 1 and y = = = 3 2 2 2 2 That is, M = 1 - 1, 32 . See Figure 10. P2 (3, 1) 5 (2) Finding the Midpoint of a Line Segment EX AMPLE 4 P1 (–5, 5) x1 + x2 y1 + y2 , ≤ 2 2 x Now Work PROBLEM r 37 Figure 10 2.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. On the real number line, the origin is assigned the number . (p. 17) 2. If - 3 and 5 are the coordinates of two points on the real number line, the distance between these points is . (pp. 19–20) 3. If 3 and 4 are the legs of a right triangle, the hypotenuse is . (p. 30) 5. The area A of a triangle whose base is b and whose altitude is h is A = . (p. 31) 6. True or False Two triangles are congruent if two angles and the included side of one equals two angles and the included side of the other. (pp. 32–33). 4. Use the converse of the Pythagorean Theorem to show that a triangle whose sides are of lengths 11, 60, and 61 is a right triangle. (pp. 30–31) Concepts and Vocabulary 7. If 1x, y2 are the coordinates of a point P in the xy-plane, then x is called the of P, and y is the of P. 8. The coordinate axes divide the xy-plane into four sections . called 9. If three distinct points P, Q, and R all lie on a line, and if d1P, Q2 = d1Q, R2, then Q is called the of the line segment from P to R. 10. True or False The distance between two points is sometimes a negative number. 11. True or False The point 1 - 1, 42 lies in quadrant IV of the Cartesian plane. 12. True or False The midpoint of a line segment is found by averaging the x-coordinates and averaging the y-coordinates of the endpoints. 13. Which of the following statements is true for a point (x, y) that lies in quadrant III? (a) Both x and y are positive. (b) Both x and y are negative. (c) x is positive, and y is negative. (d) x is negative, and y is positive. 14. Choose the formula that gives the distance between two points (x1, y1) and (x2, y2). (a) 2(x2 - x1)2 + (y2 - y1)2 (b) 2(x2 + x1)2 - (y2 + y1)2 (c) 2(x2 - x1)2 - (y2 - y1)2 (d) 2(x2 + x1)2 + (y2 + y1)2 155 SECTION 2.1 The Distance and Midpoint Formulas Skill Building In Problems 15 and 16, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies. 15. (a) A = 1 - 3, 22 (b) B = 16, 02 (c) C = 1 - 2, - 22 (d) D = 16, 52 (e) E = 10, - 32 (f) F = 16, - 32 (d) D = 14, 12 (e) E = 10, 12 (f) F = 1 - 3, 02 16. (a) A = 11, 42 (b) B = 1 - 3, - 42 (c) C = 1 - 3, 42 17. Plot the points 12, 02, 12, - 32, 12, 42, 12, 12, and 12, - 12. Describe the set of all points of the form 12, y2, where y is a real number. 18. Plot the points 10, 32, 11, 32, 1 - 2, 32, 15, 32, and 1 - 4, 32. Describe the set of all points of the form 1x, 32, where x is a real number. In Problems 19–30, find the distance d1P1 , P2 2 between the points P1 and P2 . y 19. –2 –1 y 20. 2 P = (2, 1) 2 P1 = (0, 0) 2 x 21. P2 = (–2, 1) 2 P = (0, 0) 1 –2 –1 2 x P2 (–2, 2) –2 y 2 –1 y 22. P1 = (–1, 1) 2 P1 (1, 1) 2 x –2 23. P1 = 13, - 42; P2 = 15, 42 24. P1 = 1 - 1, 02; P2 = 12, 42 25. P1 = 1 - 3, 22; P2 = 16, 02 26. P1 = 12, - 32; P2 = 14, 22 27. P1 = 14, - 32; P2 = 16, 42 28. P1 = 1 - 4, - 32; P2 = 16, 22 29. P1 = 1a, b2; P2 = 10, 02 30. P1 = 1a, a2; P2 = 10, 02 P2 = (2, 2) –1 2 x In Problems 31–36, plot each point and form the triangle ABC. Show that the triangle is a right triangle. Find its area. 31. A = 1 - 2, 52; B = 11, 32; C = 1 - 1, 02 32. A = 1 - 2, 52; B = 112, 32; C = 110, - 112 33. A = 1 - 5, 32; B = 16, 02; C = 15, 52 34. A = 1 - 6, 32; B = 13, - 52; C = 1 - 1, 52 35. A = 14, - 32; B = 10, - 32; C = 14, 22 36. A = 14, - 32; B = 14, 12; C = 12, 12 In Problems 37–44, find the midpoint of the line segment joining the points P1 and P2 . 37. P1 = 13, - 42; P2 = 15, 42 38. P1 = 1 - 2, 02; P2 = 12, 42 39. P1 = 1 - 3, 22; P2 = 16, 02 40. P1 = 12, - 32; P2 = 14, 22 41. P1 = 14, - 32; P2 = 16, 12 42. P1 = 1 - 4, - 32; P2 = 12, 22 43. P1 = 1a, b2; P2 = 10, 02 44. P1 = 1a, a2; P2 = 10, 02 Applications and Extensions 45. If the point 12, 52 is shifted 3 units to the right and 2 units down, what are its new coordinates? 46. If the point 1 - 1, 62 is shifted 2 units to the left and 4 units up, what are its new coordinates? 47. Find all points having an x-coordinate of 3 whose distance from the point 1 - 2, - 12 is 13. (a) By using the Pythagorean Theorem. (b) By using the distance formula. 48. Find all points having a y-coordinate of - 6 whose distance from the point 11, 22 is 17. (a) By using the Pythagorean Theorem. (b) By using the distance formula. 49. Find all points on the x-axis that are 6 units from the point 14, - 32 . 50. Find all points on the y-axis that are 6 units from the point 14, - 32 . (a) Find the coordinates of the point if A is shifted 3 units to the left and 4 units down. (b) Find the coordinates of the point if A is shifted 2 units to the left and 8 units up. 52. Plot the points A = 1 - 1, 82 and M = 12, 32 in the xy-plane. If M is the midpoint of a line segment AB, find the coordinates of B. 53. The midpoint of the line segment from P1 to P2 is 1 - 1, 42 . If P1 = 1 - 3, 62 , what is P2? 54. The midpoint of the line segment from P1 to P2 is 15, - 42 . If P2 = 17, - 22, what is P1? 55. Geometry The medians of a triangle are the line segments from each vertex to the midpoint of the opposite side (see the figure). Find the lengths of the medians of the triangle with vertices at A = 10, 02, B = 16, 02 , and C = 14, 42 . C 51. Suppose that A = 12, 52 are the coordinates of a point in the xy-plane. Median Midpoint A B 156 CHAPTER 2 Graphs 56. Geometry An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are 10, 42 and 10, 02, find the third vertex. How many of these s s triangles are possible? s 57. Geometry Find the midpoint of each diagonal of a square with side of length s. Draw the conclusion that the diagonals of a square intersect at their midpoints. [Hint: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the square.] a 23 a 58. Geometry Verify that the points (0, 0), (a, 0), and a , b 2 2 are the vertices of an equilateral triangle. Then show that the midpoints of the three sides are the vertices of a second equilateral triangle (refer to Problem 56). In Problems 59–62, find the length of each side of the triangle determined by the three points P1 , P2 , and P3. State whether the triangle is an isosceles triangle, a right triangle, neither of these, or both. (An isosceles triangle is one in which at least two of the sides are of equal length.) 59. P1 = 12, 12; P2 = 1 - 4, 12; P3 = 1 - 4, - 32 60. P1 = 1 - 1, 42; P2 = 16, 22; P3 = 14, - 52 61. P1 = 1 - 2, - 12; P2 = 10, 72; P3 = 13, 22 66. Little League Baseball Refer to Problem 64. Overlay a rectangular coordinate system on a Little League baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. (a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at 1180, 202, how far is it from the right fielder to second base? (c) If the center fielder is located at 1220, 2202 , how far is it from the center fielder to third base? 67. Distance between Moving Objects A Ford Focus and a Freightliner truck leave an intersection at the same time. The Focus heads east at an average speed of 30 miles per hour, while the truck heads south at an average speed of 40 miles per hour. Find an expression for their distance apart d (in miles) at the end of t hours. 68. Distance of a Moving Object from a Fixed Point A hot-air balloon, headed due east at an average speed of 15 miles per hour and at a constant altitude of 100 feet, passes over an intersection (see the figure). Find an expression for the distance d (measured in feet) from the balloon to the intersection t seconds later. 62. P1 = 17, 22; P2 = 1 - 4, 02; P3 = 14, 62 East 63. Baseball A major league baseball “diamond” is actually a square 90 feet on a side (see the figure). What is the distance directly from home plate to second base (the diagonal of the square)? 15 mph 100 ft 2nd base 90 ft 3rd base Pitching rubber 1st base 90 ft Home plate 64. Little League Baseball The layout of a Little League playing field is a square 60 feet on a side. How far is it directly from home plate to second base (the diagonal of the square)? Source: Little League Baseball, Official Regulations and Playing Rules, 2014. 65. Baseball Refer to Problem 63. Overlay a rectangular coordinate system on a major league baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. (a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at 1310, 152, how far is it from the right fielder to second base? (c) If the center fielder is located at 1300, 3002, how far is it from the center fielder to third base? 69. Drafting Error When a draftsman draws three lines that are to intersect at one point, the lines may not intersect as intended and subsequently will form an error triangle. If this error triangle is long and thin, one estimate for the location of the desired point is the midpoint of the shortest side. The figure shows one such error triangle. y (2.7, 1.7) 1.7 1.5 1.3 (2.6, 1.5) (1.4, 1.3) 1.4 2.6 2.7 x (a) Find an estimate for the desired intersection point. (b) Find the length of the median for the midpoint found in part (a). See Problem 55. 70. Net Sales The figure on page 157 illustrates how net sales of Wal-Mart Stores, Inc., grew from 2007 through 2013. Use the midpoint formula to estimate the net sales of Wal-Mart Stores, Inc., in 2010. How does your result compare to the reported value of $405 billion? Source: Wal-Mart Stores, Inc., 2013 Annual Report SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry 71. Poverty Threshold Poverty thresholds are determined by the U.S. Census Bureau. A poverty threshold represents the minimum annual household income for a family not to be considered poor. In 2003, the poverty threshold for a family of four with two children under the age of 18 years was $18,660. In 2013, the poverty threshold for a family of four with two children under the age of 18 years was $23,624. Assuming that poverty thresholds increase in a straight-line fashion, use the midpoint formula to estimate the poverty threshold for a family of four with two children under the age of 18 in 2008. How does your result compare to the actual poverty threshold in 2008 of $21,834? Source: U.S. Census Bureau Net sales ($ billions) Wal-Mart Stores, Inc. Net sales ($ billions) 500 450 400 350 300 345 250 200 150 100 50 0 2007 466 2008 2009 2010 2011 2012 157 2013 Year Explaining Concepts: Discussion and Writing 72. Write a paragraph that describes a Cartesian plane. Then write a second paragraph that describes how to plot points in the Cartesian plane. Your paragraphs should include the terms “coordinate axes,” “ordered pair,” “coordinates,” “plot,” “x-coordinate,” and “y-coordinate.” Retain Your Knowledge Problems 73–76 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 73. Determine the domain of the variable x in the expression 3x + 1 . 2x - 5 74. Find the real solution(s), if any, of the equation 3x2 - 7x - 20 = 0. 75. Multiply (7 + 3i)(1 - 2i). Write the answer in the form a + bi. 76. Solve the inequality 5(x - 3) + 2x Ú 6(2x - 3) - 7. Express the solution using interval notation. Graph the solution set. ‘Are You Prepared?’ Answers 1. 0 2. 8 3. 5 4. 112 + 602 = 121 + 3600 = 3721 = 612 5. A = 1 bh 2 6. True 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry PREPARING FOR THIS SECTION Before getting started, review the following: r Solving Linear Equations (Section 1.1, pp. 82–85) r Solve a Quadratic Equation by Factoring (Section 1.2, pp. 93–94) Now Work the ‘Are You Prepared?’ problems on page 164. OBJECTIVES 1 2 3 4 Graph Equations by Plotting Points (p. 157) Find Intercepts from a Graph (p. 159) Find Intercepts from an Equation (p. 160) Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin (p. 160) 5 Know How to Graph Key Equations (p. 163) 1 Graph Equations by Plotting Points An equation in two variables, say x and y, is a statement in which two expressions involving x and y are equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be true or false, depending on the value of the variables. Any values of x and y that result in a true statement are said to satisfy the equation. For example, the following are all equations in two variables x and y: x2 + y 2 = 5 2x - y = 6 y = 2x + 5 x2 = y 158 CHAPTER 2 Graphs The first of these, x2 + y2 = 5, is satisfied for x = 1, y = 2, since 12 + 22 = 5. Other choices of x and y, such as x = - 1, y = - 2, also satisfy this equation. It is not satisfied for x = 2 and y = 3, since 22 + 32 = 4 + 9 = 13 ≠ 5. The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates 1x, y2 satisfy the equation. Graphs play an important role in helping us to visualize the relationships that exist between two variables or quantities. Figure 11 shows the relation between the level of risk in a stock portfolio and the average annual rate of return. From the graph, we can see that when 30% of a portfolio of stocks is invested in foreign companies, risk is minimized. 18.5 Average Annual Returns (%) 18 70% 17.5 (100% foreign) 50% 40% 16 15.5 100% 60% 17 16.5 90% 80% 30% (30% foreign/70% U.S.) 20% 15 14.5 10% 0% (100% U.S.) 14 Figure 11 13.5 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18 18.5 19 19.5 20 Level of Risk (%) Source: T. Rowe Price EX AMPLE 1 Determining Whether a Point Is on the Graph of an Equation Determine if the following points are on the graph of the equation 2x - y = 6. (a) 12, 32 (b) 12, - 22 Solution (a) For the point 12, 32 , check to see whether x = 2, y = 3 satisfies the equation 2x - y = 6. 2x - y = 2122 - 3 = 4 - 3 = 1 ≠ 6 The equation is not satisfied, so the point 12, 32 is not on the graph of 2x - y = 6. (b) For the point 12, - 22 , 2x - y = 2122 - 1 - 22 = 4 + 2 = 6 The equation is satisfied, so the point 12, - 22 is on the graph of 2x - y = 6. Now Work EX AMPLE 2 PROBLEM 13 r Graphing an Equation by Plotting Points Graph the equation: y = 2x + 5 Solution y 25 (10, 25) The graph consists of all points 1x, y2 that satisfy the equation. To locate some of these points (and get an idea of the pattern of the graph), assign some numbers to x, and find corresponding values for y. If (0, 5) (1, 7) – 25 (–5, – 5) – 25 Figure 12 y = 2x + 5 25 x Then x = 0 y = 2102 + 5 = 5 x = 1 y = 2112 + 5 = 7 x = -5 y = 21 - 52 + 5 = - 5 x = 10 y = 21102 + 5 = 25 Point on Graph 10, 52 11, 72 1 - 5, - 52 110, 252 By plotting these points and then connecting them, we obtain the graph of the equation (a line), as shown in Figure 12. r SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry 159 Graphing an Equation by Plotting Points EXAMPL E 3 Graph the equation: y = x2 Solution Table 1 COMMENT Another way to obtain the graph of an equation is to use a graphing utility. Read Section 2, Using a Graphing Utility to Graph Equations, in the Appendix. ■ Graph crosses y-axis y Graph crosses x-axis Table 1 provides several points on the graph. Plotting these points and connecting them with a smooth curve gives the graph (a parabola) shown in Figure 13. x y = x2 (x, y) -4 16 ( - 4, 16) -3 9 ( - 3, 9) -2 4 ( - 2, 4) -1 1 ( - 1, 1) 0 0 (0, 0) 1 1 (1, 1) 2 4 (2, 4) 3 9 (3, 9) 4 16 (4, 16) y 20 (– 4, 16) (4, 16) 15 (–3, 9) 10 (3, 9) 5 (–2, 4) (2, 4) (1, 1) (–1, 1) (0, 0) –4 4 Figure 13 y = x 2 x r The graphs of the equations shown in Figures 12 and 13 do not show all points. For example, in Figure 12, the point 120, 452 is a part of the graph of y = 2x + 5, but it is not shown. Since the graph of y = 2x + 5 can be extended out indefinitely, we use arrows to indicate that the pattern shown continues. It is important when illustrating a graph to present enough of the graph so that any viewer of the illustration will “see” the rest of it as an obvious continuation of what is actually there. This is referred to as a complete graph. One way to obtain the complete graph of an equation is to plot enough points on the graph for a pattern to become evident. Then these points are connected with a smooth curve following the suggested pattern. But how many points are sufficient? Sometimes knowledge about the equation tells us. For example, we will learn in the next section that if an equation is of the form y = mx + b, then its graph is a line. In this case, only two points are needed to obtain the graph. One purpose of this book is to investigate the properties of equations in order to decide whether a graph is complete. Sometimes we shall graph equations by plotting points. Shortly, we shall investigate various techniques that will enable us to graph an equation without plotting so many points. Two techniques that sometimes reduce the number of points required to graph an equation involve finding intercepts and checking for symmetry. 2 Find Intercepts from a Graph x Graph touches x-axis Intercepts The points, if any, at which a graph crosses or touches the coordinate axes are called the intercepts. See Figure 14. The x-coordinate of a point at which the graph crosses or touches the x-axis is an x-intercept, and the y-coordinate of a point at which the graph crosses or touches the y-axis is a y-intercept. Figure 14 EXAMPL E 4 Find the intercepts of the graph in Figure 15. What are its x-intercepts? What are its y-intercepts? y 4 Finding Intercepts from a Graph (0, 3) Solution The intercepts of the graph are the points ( 3–2 , 0) 4 (3, 0) (0, 3.5) Figure 15 1 - 3, 02, (4.5, 0) 5 x (0, 4–3 ) 10, 32, 3 a , 0b , 2 4 a0, - b , 3 10, - 3.52, 14.5, 02 4 3 The x-intercepts are - 3, , and 4.5; the y-intercepts are - 3.5, - , and 3. 2 3 r In Example 4, note the following usage: If the type of intercept (x- versus y-) is not specified, then report the intercept as an ordered pair. However, if the type of intercept is specified, then report the coordinate of the specified intercept. For 160 CHAPTER 2 Graphs x-intercepts, report the x-coordinate of the intercept; for y-intercepts, report the y-coordinate of the intercept. Now Work 41(a) PROBLEM 3 Find Intercepts from an Equation The intercepts of a graph can be found from its equation by using the fact that points on the x-axis have y-coordinates equal to 0, and points on the y-axis have x-coordinates equal to 0. COMMENT For many equations, finding intercepts may not be so easy. In such cases, a graphing utility can be used. Read the first part of Section 3, Using a Graphing Utility to Locate Intercepts and Check for Symmetry, in the Appendix, to find out how to locate intercepts using a graphing utility. ■ EX AMPLE 5 Procedure for Finding Intercepts 1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x, where x is a real number. 2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y, where y is a real number. Finding Intercepts from an Equation Find the x-intercept(s) and the y-intercept(s) of the graph of y = x2 - 4. Then graph y = x2 - 4 by plotting points. Solution To find the x-intercept(s), let y = 0 and obtain the equation x2 - 4 1x + 22 1x - 2) x + 2 = 0 or x - 2 x = -2 or x = = = = 0 0 0 2 y = x 2 - 4 with y = 0 Factor. Zero@Product Property Solve. The equation has two solutions, - 2 and 2. The x-intercepts are - 2 and 2. To find the y-intercept(s), let x = 0 in the equation. y = x2 - 4 = 02 - 4 = - 4 The y-intercept is - 4. Since x2 Ú 0 for all x, we deduce from the equation y = x2 - 4 that y Ú - 4 for all x. This information, the intercepts, and the points from Table 2 enable us to graph y = x2 - 4. See Figure 16. Table 2 x y = x2 − 4 (x, y) (–3, 5) -3 5 -1 -3 ( - 1, - 3) 1 -3 (1, - 3) 3 5 y 5 (3, 5) ( - 3, 5) (2, 0) (– 2, 0) 5 x –5 (3, 5) (– 1, – 3) (1, – 3) (0, – 4) –5 Figure 16 y = x2 - 4 Now Work PROBLEM 23 r 4 Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin Another helpful tool for graphing equations by hand involves symmetry, particularly symmetry with respect to the x-axis, the y-axis, and the origin. Symmetry often occurs in nature. Consider the picture of the butterfly. Do you see the symmetry? SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry DEFINITION 161 A graph is said to be symmetric with respect to the x-axis if, for every point 1x, y2 on the graph, the point 1x, - y2 is also on the graph. Figure 17 illustrates the definition. Note that when a graph is symmetric with respect to the x-axis, the part of the graph above the x-axis is a reflection (or mirror image) of the part below it, and vice versa. y (x, y) (x, y ) Figure 17 Symmetry with respect to the x-axis (x, –y) (x, y ) x (x, –y ) (x, –y ) Points Symmetric with Respect to the x-Axis EXAMPL E 6 If a graph is symmetric with respect to the x-axis, and the point 13, 22 is on the graph, then the point 13, - 22 is also on the graph. r DEFINITION y (–x, y) A graph is said to be symmetric with respect to the y-axis if, for every point 1x, y2 on the graph, the point 1 - x, y2 is also on the graph. (x, y ) (–x, y) x (x, y ) Figure 18 illustrates the definition. When a graph is symmetric with respect to the y-axis, the part of the graph to the right of the y-axis is a reflection of the part to the left of it, and vice versa. Figure 18 Symmetry with respect to the y-axis Points Symmetric with Respect to the y-Axis EXAMPL E 7 If a graph is symmetric with respect to the y-axis and the point 15, 82 is on the graph, then the point 1 - 5, 82 is also on the graph. r DEFINITION y A graph is said to be symmetric with respect to the origin if, for every point 1x, y2 on the graph, the point 1 - x, - y2 is also on the graph. (x, y ) Figure 19 illustrates the definition. Symmetry with respect to the origin may be viewed in three ways: (x, y ) x (–x, –y) (–x, –y) Figure 19 Symmetry with respect to the origin EXAMPL E 8 1. As a reflection about the y-axis, followed by a reflection about the x-axis 2. As a projection along a line through the origin so that the distances from the origin are equal 3. As half of a complete revolution about the origin Points Symmetric with Respect to the Origin If a graph is symmetric with respect to the origin, and the point 14, 22 is on the graph, then the point 1 - 4, - 22 is also on the graph. Now Work PROBLEMS 31 AND 41(b) r When the graph of an equation is symmetric with respect to a coordinate axis or the origin, the number of points that you need to plot in order to see the pattern is reduced. For example, if the graph of an equation is symmetric with respect to the 162 CHAPTER 2 Graphs y-axis, then once points to the right of the y-axis are plotted, an equal number of points on the graph can be obtained by reflecting them about the y-axis. Because of this, before we graph an equation, we should first determine whether it has any symmetry. The following tests are used for this purpose. Tests for Symmetry To test the graph of an equation for symmetry with respect to the x-Axis y-Axis Origin EX AMPLE 9 Testing an Equation for Symmetry Test y = Solution Replace y by - y in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the x-axis. Replace x by - x in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the y-axis. Replace x by - x and y by −y in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the origin. 4x2 for symmetry. x2 + 1 x-Axis: To test for symmetry with respect to the x-axis, replace y by - y. Since 4x2 4x2 -y = 2 is not equivalent to y = 2 , the graph of the equation is x + 1 x + 1 not symmetric with respect to the x-axis. y-Axis: To test for symmetry with respect to the y-axis, replace x by - x. Since 41 - x2 2 4x2 4x2 , the graph of the y = = 2 is equivalent to y = 2 2 1 - x2 + 1 x + 1 x + 1 equation is symmetric with respect to the y-axis. Origin: To test for symmetry with respect to the origin, replace x by - x and y by - y. -y = 41 - x2 2 1 - x2 2 + 1 4x2 x2 + 1 4x2 y = - 2 x + 1 -y = Replace x by - x and y by - y. Simplify. Multiply both sides by - 1. Since the result is not equivalent to the original equation, the graph of the 4x2 equation y = 2 is not symmetric with respect to the origin. x + 1 r Seeing the Concept 4x2 Figure 20 shows the graph of y = 2 using a graphing utility. Do you see the symmetry with x + 1 respect to the y-axis? 5 25 Figure 20 y = 4x2 x + 1 5 25 2 Now Work PROBLEM 61 SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry 163 5 Know How to Graph Key Equations The next three examples use intercepts, symmetry, and point plotting to obtain the graphs of key equations. It is important to know the graphs of these key equations because we use them later. The first of these is y = x3. EX AM PL E 1 0 Graphing the Equation y = x 3 by Finding Intercepts, Checking for Symmetry, and Plotting Points Graph the equation y = x3 by plotting points. Find any intercepts and check for symmetry first. Solution First, find the intercepts. When x = 0, then y = 0; and when y = 0, then x = 0. The origin 10, 02 is the only intercept. Now test for symmetry. Replace y by - y. Since - y = x3 is not equivalent to y = x3, the graph is not symmetric with respect to the x-axis. Replace x by - x. Since y = 1 - x2 3 = - x3 is not equivalent to y = x3, the graph is not symmetric with respect to the y-axis. Replace x by - x and y by - y. Since - y = 1 - x2 3 = - x3 is equivalent to y = x3 (multiply both sides by - 1), the graph is symmetric with respect to the origin. x-Axis: y-Axis: y 8 (0, 0) –6 (2, 8) Origin: (1, 1) 6 (– 1, – 1) x To graph y = x3, use the equation to obtain several points on the graph. Because of the symmetry, we need to locate only points on the graph for which x Ú 0. See Table 3. Since 11, 12 is on the graph, and the graph is symmetric with respect to the origin, the point 1 - 1, - 12 is also on the graph. Plot the points from Table 3 and use the symmetry. Figure 21 shows the graph. Table 3 (– 2, – 8) –8 Figure 21 y = x3 EX AM PL E 11 x y = x3 (x, y) 0 0 (0, 0) 1 1 (1, 1) 2 8 (2, 8) 3 27 (3, 27) Graphing the Equation x = y r 2 (a) Graph the equation x = y2. Find any intercepts and check for symmetry first. (b) Graph x = y2, y Ú 0. Solution (a) The lone intercept is 10, 02 . The graph is symmetric with respect to the x-axis. (Do you see why? Replace y by - y.) Figure 22 shows the graph. (b) If we restrict y so that y Ú 0, the equation x = y2, y Ú 0, may be written equivalently as y = 1x. The portion of the graph of x = y2 in quadrant I is therefore the graph of y = 1x. See Figure 23. y 6 y 6 (9, 3) (1, 1) (0, 0) 6 Y1 5 √x 22 –2 (1, – 1) 10 26 Figure 24 Y2 5 2√x (4, 2) 5 (1, 1) 10 x (4, 2) (9, 3) (0, 0) (4, – 2) (9, – 3) Figure 22 x = y2 –2 5 Figure 23 y = 1x 10 x r COMMENT To see the graph of the equation x = y 2 on a graphing calculator, you will need to graph two equations: Y1 = 1x and Y2 = - 1x. We discuss why in Chapter 3. See Figure 24. ■ 164 CHAPTER 2 Graphs EX A MPL E 12 Graphing the Equation y = 1 x 1 . First, find any intercepts and check for symmetry. x Check for intercepts first. If we let x = 0, we obtain 0 in the denominator, which makes y undefined. We conclude that there is no y-intercept. If we let y = 0, we get 1 the equation = 0, which has no solution. We conclude that there is no x-intercept. x 1 The graph of y = does not cross or touch the coordinate axes. x Next check for symmetry: 1 1 x-Axis: Replacing y by - y yields - y = , which is not equivalent to y = . x x 1 1 y-Axis: Replacing x by - x yields y = = - , which is not equivalent to -x x 1 y = . x 1 Origin: Replacing x by - x and y by - y yields - y = - , which is equivalent to x 1 y = . The graph is symmetric with respect to the origin. x Graph the equation y = Solution Table 4 1 x x y = 1 10 10 1 3 3 1 a , 3b 3 1 2 2 1 a , 2b 2 1 1 (1, 1) 2 1 2 1 a2, b 2 3 1 3 1 a3, b 3 10 1 10 (x, y) a 1 , 10b 10 a10, 1 b 10 y 3 (––12 , 2) (1, 1) –3 (2, ––12 ) 3 (–2, – ––12 ) (–1, –1) x Now set up Table 4, listing several points on the graph. Because of the symmetry with respect to the origin, we use only positive values of x. From Table 4 we 1 infer that if x is a large and positive number, then y = is a positive number close x 1 to 0. We also infer that if x is a positive number close to 0, then y = is a large and x positive number. Armed with this information, we can graph the equation. 1 Figure 25 illustrates some of these points and the graph of y = . Observe how x the absence of intercepts and the existence of symmetry with respect to the origin were utilized. r 1 COMMENT Refer to Example 2 in the Appendix, Section 3, for the graph of y = found x using a graphing utility. ■ (– ––12 , –2) –3 Figure 25 y = 1 x 2.2 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Solve the equation 21x + 32 - 1 = - 7. (pp. 82–85) 2. Solve the equation x2 - 9 = 0. (pp. 93–94) Concepts and Vocabulary 3. The points, if any, at which a graph crosses or touches the coordinate axes are called . 8. True or False To find the y-intercepts of the graph of an equation, let x = 0 and solve for y. 4. The x-intercepts of the graph of an equation are those . x-values for which 9. True or False The y-coordinate of a point at which the graph crosses or touches the x-axis is an x-intercept. 5. If for every point 1x, y2 on the graph of an equation the point 1 - x, y2 is also on the graph, then the graph is symmetric with respect to the . 6. If the graph of an equation is symmetric with respect to the is also y-axis and - 4 is an x-intercept of this graph, then an x-intercept. 7. If the graph of an equation is symmetric with respect to the origin and 13, - 42 is a point on the graph, then is also a point on the graph. 10. True or False If a graph is symmetric with respect to the x-axis, then it cannot be symmetric with respect to the y-axis. 11. Given that the intercepts of a graph are (–4, 0) and (0, 5), choose the statement that is true. (a) The y-intercept is –4, and the x-intercept is 5. (b) The y-intercepts are –4 and 5. (c) The x-intercepts are –4 and 5. (d) The x-intercept is –4, and the y-intercept is 5. SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry 12. To test whether the graph of an equation is symmetric with respect to the origin, replace in the equation and simplify. If an equivalent equation results, then the graph is symmetric with respect to the origin. (a) (b) (c) (d) 165 x by –x y by –y x by –x and y by –y x by –y and y by –x Skill Building In Problems 13–18, determine which of the given points are on the graph of the equation. 13. Equation: y = x4 - 1x Points: 10, 02; 11, 12; 12, 42 14. Equation: y = x3 - 21x Points: 10, 02; 11, 12; 11, - 12 15. Equation: y2 = x2 + 9 Points: 10, 32; 13, 02; 1 - 3, 02 16. Equation: y3 = x + 1 Points: 11, 22; 10, 12; 1 - 1, 02 17. Equation: x2 + y2 = 4 Points: 10, 22; 1 - 2, 22; 18. Equation: x2 + 4y2 = 4 1 22, 22 2 1 Points: 10, 12; 12, 02; a2, b 2 In Problems 19–30, find the intercepts and graph each equation by plotting points. Be sure to label the intercepts. 19. y = x + 2 20. y = x - 6 21. y = 2x + 8 22. y = 3x - 9 23. y = x - 1 24. y = x - 9 25. y = - x + 4 26. y = - x2 + 1 27. 2x + 3y = 6 28. 5x + 2y = 10 29. 9x2 + 4y = 36 30. 4x2 + y = 4 2 2 2 In Problems 31–40, plot each point. Then plot the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin. 31. 13, 42 32. 15, 32 36. 1 - 1, - 12 33. 1 - 2, 12 37. 1 - 3, - 42 34. 14, - 22 38. 14, 02 35. 15, - 22 39. 10, - 32 40. 1 - 3, 02 In Problems 41–52, the graph of an equation is given. (a) Find the intercepts. (b) Indicate whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin. 41. 42. y 3 46. y 3 3x –3 –– 3x 2 1 x –– 2 3 47. 50. 3x 3 3 y 40 x 3x 3 3x –3 48. y 6 6 6 6 –3 y 3 3 x 4 y 3 –3 49. y 4 3 –3 45. 44. y 1 3 3x –3 43. y 3 40 51. y 3 52. 8 −2 3x 3 4 3 2 −4 4 −4 −8 In Problems 53–56, draw a complete graph so that it has the type of symmetry indicated. 53. y-axis 54. x-axis y 9 y 55. Origin y 5 4 (0, 2) (5, 3) (0, 0) (–4, 0) 5x –5 9x (2, –5) –9 –9 (0, –9) 56. y-axis , 2) (–– 2 y (0, 4) 4 (2, 2) (, 0) x (0, 0) 3x –3 –5 –4 –2 166 CHAPTER 2 Graphs In Problems 57–72, list the intercepts and test for symmetry. 57. y2 = x + 4 58. y2 = x + 9 3 59. y = 2 x 5 60. y = 2 x 61. x2 + y - 9 = 0 62. x2 - y - 4 = 0 63. 9x2 + 4y2 = 36 64. 4x2 + y2 = 4 65. y = x3 - 27 66. y = x4 - 1 67. y = x2 - 3x - 4 68. y = x2 + 4 69. y = 3x x + 9 2 70. y = x2 - 4 2x 71. y = - x3 x - 9 2 72. y = x4 + 1 2x5 In Problems 73–76, draw a quick sketch of the graph of each equation. 73. y = x3 74. x = y2 77. If 1a, 42 is a point on the graph of y = x2 + 3x, what is a? 75. y = 1x 76. y = 1 x 78. If 1a, - 52 is a point on the graph of y = x2 + 6x, what is a? Applications and Extensions 79. Given that the point (1, 2) is on the graph of an equation that is symmetric with respect to the origin, what other point is on the graph? 80. If the graph of an equation is symmetric with respect to the y-axis and 6 is an x-intercept of this graph, name another x-intercept. 81. If the graph of an equation is symmetric with respect to the origin and - 4 is an x-intercept of this graph, name another x-intercept. 82. If the graph of an equation is symmetric with respect to the x-axis and 2 is a y-intercept, name another y-intercept. 83. Microphones In studios and on stages, cardioid microphones are often preferred for the richness they add to voices and for their ability to reduce the level of sound from the sides and rear of the microphone. Suppose one such cardioid pattern is given by the equation 1x2 + y2 - x2 2 = x2 + y2. (a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, the y-axis, and the origin. Source: www.notaviva.com 84. Solar Energy The solar electric generating systems at Kramer Junction, California, use parabolic troughs to heat a heat-transfer fluid to a high temperature. This fluid is used to generate steam that drives a power conversion system to produce electricity. For troughs 7.5 feet wide, an equation for the cross section is 16y2 = 120x - 225. (a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, the y-axis, and the origin. Source: U.S. Department of Energy Explaining Concepts: Discussion and Writing 85. (a) Graph y = 2x2 , y = x, y = 0 x 0 , and y = 1 1x2 2, noting which graphs are the same. (b) Explain why the graphs of y = 2x2 and y = 0 x 0 are the same. (c) Explain why the graphs of y = x and y = 1 1x2 2 are not the same. (d) Explain why the graphs of y = 2x2 and y = x are not the same. 86. Explain what is meant by a complete graph. 87. Draw a graph of an equation that contains two x-intercepts; at one the graph crosses the x-axis, and at the other the graph touches the x-axis. 88. Make up an equation with the intercepts 12, 02, 14, 02 , and 10, 12 . Compare your equation with a friend’s equation. Comment on any similarities. 89. Draw a graph that contains the points 1 - 2, - 12, 10, 12, 11, 32 , and 13, 52 . Compare your graph with those of other students. Are most of the graphs almost straight lines? How many are “curved”? Discuss the various ways in which these points might be connected. 90. An equation is being tested for symmetry with respect to the x-axis, the y-axis, and the origin. Explain why, if two of these symmetries are present, the remaining one must also be present. 91. Draw a graph that contains the points ( - 2, 5), ( - 1, 3), and (0, 2) and is symmetric with respect to the y-axis. Compare your graph with those of other students; comment on any similarities. Can a graph contain these points and be symmetric with respect to the x-axis? the origin? Why or why not? SECTION 2.3 Lines 167 Retain Your Knowledge Problems 92–95 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x + y 94. Simplify: 2- 196 if x = 6 and y = - 2. 92. Find the value of x - y 95. Solve x2 - 8x + 4 = 0 by completing the square. 2 93. Factor 3x - 30x + 75 completely. ‘Are You Prepared?’ Answers 1. 5 - 66 2. 5 - 3, 36 2.3 Lines OBJECTIVES 1 2 3 4 5 6 7 8 9 10 Calculate and Interpret the Slope of a Line (p. 167) Graph Lines Given a Point and the Slope (p. 170) Find the Equation of a Vertical Line (p. 170) Use the Point–Slope Form of a Line; Identify Horizontal Lines (p. 171) Find the Equation of a Line Given Two Points (p. 172) Write the Equation of a Line in Slope–Intercept Form (p. 172) Identify the Slope and y-Intercept of a Line from Its Equation (p. 173) Graph Lines Written in General Form Using Intercepts (p. 174) Find Equations of Parallel Lines (p. 175) Find Equations of Perpendicular Lines (p. 176) In this section we study a certain type of equation that contains two variables, called a linear equation, and its graph, a line. Line Rise Run Figure 26 DEFINITION 1 Calculate and Interpret the Slope of a Line Consider the staircase illustrated in Figure 26. Each step contains exactly the same horizontal run and the same vertical rise. The ratio of the rise to the run, called the slope, is a numerical measure of the steepness of the staircase. For example, if the run is increased and the rise remains the same, the staircase becomes less steep. If the run is kept the same but the rise is increased, the staircase becomes more steep. This important characteristic of a line is best defined using rectangular coordinates. Let P = 1x1 , y1 2 and Q = 1x2 , y2 2 be two distinct points. If x1 ≠ x2 , the slope m of the nonvertical line L containing P and Q is defined by the formula m = y2 - y1 x2 - x1 x1 ≠ x2 (1) If x1 = x2 , then L is a vertical line and the slope m of L is undefined (since this results in division by 0). Figure 27(a) on page 168 provides an illustration of the slope of a nonvertical line; Figure 27(b) illustrates a vertical line. 168 CHAPTER 2 Graphs L L Q = (x 2, y2) y2 Run = x2 – x1 x1 Figure 27 Q = (x 1, y2) y1 P = (x 1, y1) Rise = y2 – y1 P = (x 1, y1) y1 y2 x2 (a) Slope of L is m = x1 y2 – y1 _______ , x Z x x2 – x1 1 2 (b) Slope is undefined; L is vertical As Figure 27(a) illustrates, the slope m of a nonvertical line may be viewed as In Words The symbol ∆ is the Greek uppercase letter delta. In mathematics, ∆ ∆y is read “change in,” so is read ∆x “change in y divided by change in x.” m = y2 - y1 Rise = x2 - x1 Run or as m = Change in y y2 - y1 ∆y = = x2 - x1 Change in x ∆x That is, the slope m of a nonvertical line measures the amount y changes when x ∆y is called the average rate of change of y changes from x1 to x2. The expression ∆x with respect to x. Two comments about computing the slope of a nonvertical line may prove helpful: 1. Any two distinct points on the line can be used to compute the slope of the line. (See Figure 28 for justification.) Since any two distinct points can be used to compute the slope of a line, the average rate of change of a line is always the same number. y Figure 28 Triangles ABC and PQR are similar (equal angles), so ratios of corresponding sides are proportional. Then y2 - y1 Slope using P and Q = = x2 - x1 d(B, C) = Slope using A and B d(A, C) Q = (x 2, y2) y2 – y1 P = (x 1, y1) B x2 – x1 R x A C 2. The slope of a line may be computed from P = 1x1 , y1 2 to Q = 1x2 , y2 2 or from Q to P because y2 - y1 y1 - y2 = x2 - x1 x1 - x2 EX AMPLE 1 Finding and Interpreting the Slope of a Line Given Two Points The slope m of the line containing the points 11, 22 and 15, - 32 may be computed as m = -3 - 2 -5 5 = = 5 - 1 4 4 or as m = 2 - 1 - 32 5 5 = = 1 - 5 -4 4 For every 4-unit change in x, y will change by - 5 units. That is, if x increases by 4 units, then y will decrease by 5 units. The average rate of change of y with respect 5 to x is - . 4 r Now Work PROBLEMS 13 AND 19 SECTION 2.3 Lines 169 Finding the Slopes of Various Lines Containing the Same Point (2, 3) EXAMPL E 2 Compute the slopes of the lines L1 , L2 , L3 , and L4 containing the following pairs of points. Graph all four lines on the same set of coordinate axes. L1 : L2 : L3 : L4 : Solution P P P P = = = = 12, 32 12, 32 12, 32 12, 32 Q1 Q2 Q3 Q4 = = = = 1 - 1, - 22 13, - 12 15, 32 12, 52 Let m1 , m2 , m3 , and m4 denote the slopes of the lines L1 , L2, L3 , and L4 , respectively. Then -2 - 3 -5 5 A rise of 5 divided by a run of 3 = = -1 - 2 -3 3 -1 - 3 -4 m2 = = = -4 3 - 2 1 0 3 - 3 = = 0 m3 = 5 - 2 3 m4 is undefined because x1 = x2 = 2 m1 = L2 L4 L1 y 5 Q4 5 (2, 5) P 5 (2, 3) m3 5 0 L3 Q3 5 (5, 3) 25 m1 5 5 3 The graphs of these lines are given in Figure 29. 5 x Q2 5 (3, 21) Q1 5 (21, 22) 23 m4 undefined r Figure 29 illustrates the following facts: m2 5 24 1. When the slope of a line is positive, the line slants upward from left to right 1L1 2 . 2. When the slope of a line is negative, the line slants downward from left to right 1L2 2 . 3. When the slope is 0, the line is horizontal 1L3 2 . 4. When the slope is undefined, the line is vertical 1L4 2 . Figure 29 Seeing the Concept Y6 5 6x Y5 5 2x Y4 5 x Y3 5 2 On the same screen, graph the following equations: Y1 = 0 1 Y2 = x 4 1 Y3 = x 2 Y4 = x Y5 = 2x Y6 = 6x 1 x 2 1 Y2 5 4 x 23 3 Y1 5 0 22 Figure 30 Slope of line is 0. 1 Slope of line is . 4 1 Slope of line is . 2 Slope of line is 1. Slope of line is 2. Slope of line is 6. See Figure 30. Seeing the Concept Y4 5 2x Y6 5 26x Y5 5 22x On the same screen, graph the following equations: Y1 = 0 1 Y3 5 2 2 x 2 1 Y2 5 2 4 x 23 Slope of line is - Y3 = Slope of line is - Y4 = Y5 = Y6 = 22 Figure 31 Y2 = - 3 Y1 5 0 See Figure 31. 1 x 4 1 - x 2 -x - 2x - 6x Slope of line is 0. 1 . 4 1 . 2 Slope of line is - 1. Slope of line is - 2. Slope of line is - 6. 170 CHAPTER 2 Graphs Figures 30 and 31 on page 169 illustrate that the closer the line is to the vertical position, the greater the magnitude of the slope. 2 Graph Lines Given a Point and the Slope EX AMPLE 3 Graphing a Line Given a Point and a Slope Draw a graph of the line that contains the point 13, 22 and has a slope of: (a) Solution y 6 (7, 5) Rise = 3 (3, 2) Run = 4 (b) - Rise 3 . The slope means that for every horizontal movement (run) Run 4 of 4 units to the right, there will be a vertical movement (rise) of 3 units. Start at the given point 13, 22 and move 4 units to the right and 3 units up, arriving at the point 17, 52 . Drawing the line through this point and the point 13, 22 gives the graph. See Figure 32. (b) The fact that the slope is - y (–2, 6) 6 Rise = 4 4 -4 Rise = = 5 5 Run means that for every horizontal movement of 5 units to the right, there will be a corresponding vertical movement of - 4 units (a downward movement). Start at the given point 13, 22 and move 5 units to the right and then 4 units down, arriving at the point 18, - 22 . Drawing the line through these points gives the graph. See Figure 33. Alternatively, consider that Figure 32 (3, 2) Run = 5 Run = –5 - Rise = –4 10 x –2 –2 4 5 (a) Slope = 10 x 5 –2 3 4 4 4 Rise = = 5 -5 Run so for every horizontal movement of - 5 units (a movement to the left), there will be a corresponding vertical movement of 4 units (upward). This approach leads to the point 1 - 2, 62 , which is also on the graph of the line in Figure 33. (8, –2) Figure 33 Now Work PROBLEM 25 (graph the line) r 3 Find the Equation of a Vertical Line EX AMPLE 4 Graphing a Line Graph the equation: x = 3 Solution To graph x = 3, we find all points 1x, y2 in the plane for which x = 3. No matter what y-coordinate is used, the corresponding x-coordinate always equals 3. Consequently, the graph of the equation x = 3 is a vertical line with x-intercept 3 and undefined slope. See Figure 34. y 4 (3, 3) (3, 2) (3, 1) 1 1 (3, 0) (3, 1) Figure 34 x = 3 5 x r SECTION 2.3 Lines 171 Example 4 suggests the following result: THEOREM Equation of a Vertical Line A vertical line is given by an equation of the form x = a where a is the x-intercept. COMMENT To graph an equation using a graphing utility, we need to express the equation in the form y = 5 expression in x6 . But x = 3 cannot be put in this form. To overcome this, most graphing utilities have special commands for drawing vertical lines. DRAW, LINE, PLOT, and VERT are among the more common ones. Consult your manual to determine the correct methodology for your graphing utility. ■ y L (x, y ) Let L be a nonvertical line with slope m that contains the point 1x1 , y1 2. See Figure 35. For any other point 1x, y2 on L, we have y – y1 (x 1, y1) 4 Use the Point–Slope Form of a Line; Identify Horizontal Lines x – x1 m = x y - y1 x - x1 or y - y1 = m1x - x1 2 Figure 35 THEOREM Point–Slope Form of an Equation of a Line An equation of a nonvertical line with slope m that contains the point 1x1 , y1 2 is y - y1 = m1x - x1 2 EXAMPL E 5 y 6 (2, 6) Using the Point–Slope Form of a Line An equation of the line with slope 4 that contains the point 11, 22 can be found by using the point–slope form with m = 4, x1 = 1, and y1 = 2. y - y1 = m1x - x1 2 Rise 5 4 (1, 2) y - 2 = 41x - 12 Run 5 1 22 y = 4x - 2 5 (2) m = 4, x1 = 1, y1 = 2 Solve for y. x See Figure 36 for the graph. Now Work PROBLEM 25 (find the point-slope form) r Figure 36 y = 4x - 2 EXAMPL E 6 Solution y (3, 2) 1 3 Figure 37 y = 2 Find an equation of the horizontal line containing the point 13, 22 . Because all the y-values are equal on a horizontal line, the slope of a horizontal line is 0. To get an equation, use the point–slope form with m = 0, x1 = 3, and y1 = 2. y - y1 = m1x - x1 2 4 –1 Finding the Equation of a Horizontal Line 5 x y - 2 = 0 # 1x - 32 m = 0, x1 = 3, and y1 = 2 y - 2 = 0 y = 2 See Figure 37 for the graph. r 172 CHAPTER 2 Graphs Example 6 suggests the following result: THEOREM Equation of a Horizontal Line A horizontal line is given by an equation of the form y = b where b is the y-intercept. 5 Find the Equation of a Line Given Two Points EX A MPL E 7 Solution Finding an Equation of a Line Given Two Points Find an equation of the line containing the points 12, 32 and 1 - 4, 52 . Graph the line. First compute the slope of the line. m = 2 1 5 - 3 = = -4 - 2 -6 3 (–4, 5) (2, 3) 2 –2 y2 - y1 x2 - x1 1 Use the point 12, 32 and the slope m = - to get the point–slope form of the 3 equation of the line. y –4 m = 10 x 1 Figure 38 y - 3 = - (x - 2) 3 y - 3 = - 1 1x - 22 3 y - y1 = m(x - x1) See Figure 38 for the graph. r In the solution to Example 7, we could have used the other point, 1 - 4, 52 , instead of the point 12, 32 . The equation that results, although it looks different, is equivalent to the equation that we obtained in the example. (Try it for yourself.) Now Work PROBLEM 39 6 Write the Equation of a Line in Slope–Intercept Form Another useful equation of a line is obtained when the slope m and y-intercept b are known. In this event, both the slope m of the line and a point 10, b2 on the line are known; then use the point–slope form, equation (2), to obtain the following equation: y - b = m1x - 02 THEOREM or y = mx + b Slope–Intercept Form of an Equation of a Line An equation of a line with slope m and y-intercept b is y = mx + b Now Work PROBLEMS 47 AND 53 (express answer in slope–intercept form) (3) SECTION 2.3 Lines Y5 5 23x 1 2 Y4 5 3x 1 2 Seeing the Concept Y2 5x 1 2 To see the role that the slope m plays, graph the following lines on the same screen. 4 Y3 52x 1 2 Y1 5 2 26 173 6 Y1 = 2 m = 0 Y2 = x + 2 m = 1 Y3 = - x + 2 m = -1 Y4 = 3x + 2 m = 3 Y5 = - 3x + 2 m = - 3 See Figure 39. What do you conclude about the lines y = mx + 2? 24 Figure 39 y = mx + 2 Y2 5 2x 1 1 Y1 5 2x Y3 5 2x 2 1 Seeing the Concept To see the role of the y-intercept b, graph the following lines on the same screen. Y1 = 2x 4 b = 0 Y2 = 2x + 1 b = 1 Y4 5 2x 1 4 Y3 = 2x - 1 b = - 1 26 Y4 = 2x + 4 b = 4 6 24 Y5 = 2x - 4 b = - 4 Y5 5 2x 2 4 See Figure 40. What do you conclude about the lines y = 2x + b? Figure 40 y = 2x + b 7 Identify the Slope and y-Intercept of a Line from Its Equation When the equation of a line is written in slope–intercept form, it is easy to find the slope m and y-intercept b of the line. For example, suppose that the equation of a line is y = - 2x + 7 Compare this equation to y = mx + b. y = - 2x + 7 y = c c mx + b The slope of this line is - 2 and its y-intercept is 7. Now Work EXAMPL E 8 PROBLEM 73 Finding the Slope and y-Intercept Find the slope m and y-intercept b of the equation 2x + 4y = 8. Graph the equation. Solution To obtain the slope and y-intercept, write the equation in slope–intercept form by solving for y. 2x + 4y = 8 4y = - 2x + 8 y = - y 4 (0, 2) –3 2 1 (2, 1) 3 1 Figure 41 y = - x + 2 2 x 1 x + 2 2 y = mx + b 1 The coefficient of x, - , is the slope, and the constant, 2, is the y-intercept. Graph 2 1 the line with y-intercept 2 and with slope - . Starting at the point 10, 22, go to the 2 right 2 units and then down 1 unit to the point 12, 12 . Draw the line through these points. See Figure 41. Now Work PROBLEM 79 r 174 CHAPTER 2 Graphs 8 Graph Lines Written in General Form Using Intercepts Refer to Example 8. The form of the equation of the line 2x + 4y = 8 is called the general form. DEFINITION The equation of a line is in general form* when it is written as Ax + By = C (4) where A, B, and C are real numbers and A and B are not both 0. If B = 0 in equation (4), then A ≠ 0 and the graph of the equation is a vertical C line: x = . If B ≠ 0 in equation (4), then we can solve the equation for y and A write the equation in slope–intercept form as we did in Example 8. Another approach to graphing equation (4) is to find its intercepts. Remember, the intercepts of the graph of an equation are the points where the graph crosses or touches a coordinate axis. EX AMPLE 9 Graphing an Equation in General Form Using Its Intercepts Graph the equation 2x + 4y = 8 by finding its intercepts. Solution To obtain the x-intercept, let y = 0 in the equation and solve for x. 2x + 4y = 8 2x + 4102 = 8 Let y = 0. 2x = 8 x = 4 Divide both sides by 2. The x-intercept is 4, and the point 14, 02 is on the graph of the equation. To obtain the y-intercept, let x = 0 in the equation and solve for y. 2x + 4y = 8 y 4 2102 + 4y = 8 4y = 8 (0, 2) y = 2 (4, 0) –3 Let x = 0. 3 Figure 42 2x + 4y = 8 x Divide both sides by 4. The y-intercept is 2, and the point 10, 22 is on the graph of the equation. Plot the points 14, 02 and 10, 22 and draw the line through the points. See Figure 42. Now Work PROBLEM r 93 Every line has an equation that is equivalent to an equation written in general form. For example, a vertical line whose equation is x = a can be written in the general form 1#x + 0#y = a A = 1, B = 0, C = a A horizontal line whose equation is y = b can be written in the general form 0#x + 1#y = b *Some texts use the term standard form. A = 0, B = 1, C = b SECTION 2.3 Lines 175 Lines that are neither vertical nor horizontal have general equations of the form Ax + By = C A ≠ 0 and B ≠ 0 Because the equation of every line can be written in general form, any equation equivalent to equation (4) is called a linear equation. 9 Find Equations of Parallel Lines y Rise Rise When two lines (in the plane) do not intersect (that is, they have no points in common), they are parallel. Look at Figure 43. There we have drawn two parallel lines and have constructed two right triangles by drawing sides parallel to the coordinate axes. The right triangles are similar. (Do you see why? Two angles are equal.) Because the triangles are similar, the ratios of corresponding sides are equal. Run Run x THEOREM Criteria for Parallel Lines Two nonvertical lines are parallel if and only if their slopes are equal and they have different y-intercepts. Figure 43 Parallel lines The use of the phrase “if and only if” in the preceding theorem means that actually two statements are being made, one the converse of the other. If two nonvertical lines are parallel, then their slopes are equal and they have different y-intercepts. If two nonvertical lines have equal slopes and they have different y-intercepts, then they are parallel. EX AM PL E 1 0 Showing That Two Lines Are Parallel Show that the lines given by the following equations are parallel. L1 : 2x + 3y = 6 Solution To determine whether these lines have equal slopes and different y-intercepts, write each equation in slope–intercept form. L1 : 2x + 3y = 6 y 5 L2 : 4x + 6y = 0 3y = - 2x + 6 y = - 5 L2 : 4x + 6y = 0 5 x L1 L2 Figure 44 Slope = - 2 ; 3 6y = - 4x 2 x + 2 3 y@intercept = 2 y = Slope = - 2 ; 3 2 x 3 y@intercept = 0 2 Because these lines have the same slope, - , but different y-intercepts, the lines are 3 parallel. See Figure 44. r EX AM PL E 1 1 Solution Finding a Line That Is Parallel to a Given Line Find an equation for the line that contains the point 12, - 32 and is parallel to the line 2x + y = 6. Since the two lines are to be parallel, the slope of the line being sought equals the slope of the line 2x + y = 6. Begin by writing the equation of the line 2x + y = 6 in slope–intercept form. 2x + y = 6 y = - 2x + 6 176 CHAPTER 2 Graphs The slope is - 2. Since the line being sought also has slope - 2 and contains the point 12, - 32 , use the point–slope form to obtain its equation. y y - y1 = m1x - x1 2 6 y - 1 - 32 = - 21x - 22 y + 3 = - 2x + 4 6 y = - 2x + 1 6 x 2x y 6 (2, 3) 5 2x y 1 Figure 45 2x + y = 1 Point–slope form m = - 2, x1 = 2, y1 = - 3 Simplify. Slope–intercept form General form This line is parallel to the line 2x + y = 6 and contains the point 12, - 32 . See Figure 45. Now Work PROBLEM r 61 y 10 Find Equations of Perpendicular Lines When two lines intersect at a right angle (90°), they are perpendicular. See Figure 46. The following result gives a condition, in terms of their slopes, for two lines to be perpendicular. 90° x Figure 46 Perpendicular lines THEOREM Criterion for Perpendicular Lines Two nonvertical lines are perpendicular if and only if the product of their slopes is - 1. Here we shall prove the “only if” part of the statement: If two nonvertical lines are perpendicular, then the product of their slopes is - 1. In Problem 130 you are asked to prove the “if” part of the theorem: If two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular. y Slope m2 A = (1, m2) Slope m1 Rise = m 2 x Run = 1 O 1 3 d 1O, A2 4 2 + 3 d 1O, B2 4 2 = 3 d 1A, B2 4 2 (5) Rise = m1 B = (1, m1) Figure 47 Proof Let m1 and m2 denote the slopes of the two lines. There is no loss in generality (that is, neither the angle nor the slopes are affected) if we situate the lines so that they meet at the origin. See Figure 47. The point A = 11, m2 2 is on the line having slope m2 , and the point B = 11, m1 2 is on the line having slope m1 . (Do you see why this must be true?) Suppose that the lines are perpendicular. Then triangle OAB is a right triangle. As a result of the Pythagorean Theorem, it follows that Using the distance formula, the squares of these distances are 3 d 1O, A2 4 2 = 11 - 02 2 + 1m2 - 02 2 = 1 + m22 3 d 1O, B2 4 2 = 11 - 02 2 + 1m1 - 02 2 = 1 + m21 3 d 1A, B2 4 2 = 11 - 12 2 + 1m2 - m1 2 2 = m22 - 2m1 m2 + m21 SECTION 2.3 Lines 177 Using these facts in equation (5), we get 11 + m22 2 + 11 + m21 2 = m22 - 2m1 m2 + m21 which, upon simplification, can be written as m1 m2 = - 1 If the lines are perpendicular, the product of their slopes is - 1. ■ You may find it easier to remember the condition for two nonvertical lines to be perpendicular by observing that the equality m1 m2 = - 1 means that m1 and m2 are 1 1 negative reciprocals of each other; that is, m1 = and m2 = . m2 m1 Finding the Slope of a Line Perpendicular to Another Line EX AM PL E 1 2 3 2 If a line has slope , any line having slope - is perpendicular to it. 2 3 r Finding the Equation of a Line Perpendicular to a Given Line EX AM PL E 1 3 Find an equation of the line that contains the point 11, - 22 and is perpendicular to the line x + 3y = 6. Graph the two lines. Solution First write the equation of the given line in slope–intercept form to find its slope. x + 3y = 6 3y = - x + 6 y = - 1 x + 2 3 Proceed to solve for y. Place in the form y = mx + b. 1 The given line has slope - . Any line perpendicular to this line will have slope 3. 3 Because the point 11, - 22 is on this line with slope 3, use the point–slope form of the equation of a line. y y 3x 5 6 x 3y 6 y - y1 = m1x - x1 2 y - 1 - 22 = 31x - 12 4 2 2 2 4 6 To obtain other forms of the equation, proceed as follows: (1, 2) y + 2 = 3x - 3 y = 3x - 5 4 3x - y = 5 Figure 48 m = 3, x1 = 1, y1 = - 2 y + 2 = 31x - 12 x 2 Point–slope form Figure 48 shows the graphs. Now Work PROBLEM Simplify. Slope–intercept form General form r 67 WARNING Be sure to use a square screen when you use a graphing calculator to graph perpendicular lines. Otherwise, the angle between the two lines will appear distorted. A discussion of square screens is given in Section 5 of the Appendix. ■ 178 CHAPTER 2 Graphs 2.3 Assess Your Understanding Concepts and Vocabulary 1. The slope of a vertical line is horizontal line is . ; the slope of a 2. For the line 2x + 3y = 6, the x-intercept is y-intercept is . and the 3. True or False The equation 3x + 4y = 6 is written in general form. 4. True or False The slope of the line 2y = 3x + 5 is 3. 5. True or False The point 11, 22 is on the line 2x + y = 4. 6. Two nonvertical lines have slopes m1 and m2 , respectively. The lines are parallel if and the are unequal; the lines are perpendicular if . 7. The lines y = 2x + 3 and y = ax + 5 are parallel if . a = 8. The lines y = 2x - 1 and y = ax + 2 are perpendicular if a = . 9. True or False Perpendicular lines have slopes that are reciprocals of one another. 10. Choose the formula for finding the slope m of a nonvertical line that contains the two distinct points (x1, y1) and (x2, y2). y2 - x2 (a) m = x1 ≠ y1 y1 - x1 y2 - x1 (b) m = y1 ≠ x2 x2 - y1 x2 - x1 (c) m = y1 ≠ y2 y2 - y1 y2 - y1 (d) m = x1 ≠ x2 x2 - x1 11. If a line slants downward from left to right, then which of the following describes its slope? (a) positive (b) zero (c) negative (d) undefined 12. Choose the correct statement about the graph of the line y = - 3. (a) The graph is vertical with x-intercept - 3. (b) The graph is horizontal with y-intercept - 3. (c) The graph is vertical with y-intercept - 3. (d) The graph is horizontal with x-intercept - 3. Skill Building In Problems 13–16, (a) find the slope of the line and (b) interpret the slope. 13. 14. y 2 (–2, 1) 2 (2, 1) 16. y (–2, 2) 2 (1, 1) y (–1, 1) 2 (2, 2) (0, 0) (0, 0) –2 15. y 2 –1 x –2 x 2 –1 –2 2 –1 x –2 x 2 –1 In Problems 17–24, plot each pair of points and determine the slope of the line containing them. Graph the line. 17. 12, 32; 14, 02 21. 1 - 3, - 12; 12, - 12 18. 14, 22; 13, 42 19. 1 - 2, 32; 12, 12 22. 14, 22; 1 - 5, 22 23. 1 - 1, 22; 1 - 1, - 22 20. 1 - 1, 12; 12, 32 24. 12, 02; 12, 22 In Problems 25–32, graph the line that contains the point P and has slope m. In Problems 25–30, find the point-slope form of the equation of the line. In Problems 31 and 32, find an equation of the line. 25. P = 11, 22; m = 3 26. P = 12, 12; m = 4 29. P = 1 - 1, 32; m = 0 30. P = 12, - 42; m = 0 27. P = 12, 42; m = - 3 4 31. P = 10, 32 ; slope undefined 28. P = 11, 32; m = - 2 5 32. P = 1 - 2, 02 ; slope undefined In Problems 33–38, the slope and a point on a line are given. Use this information to locate three additional points on the line. Answers may vary. [Hint: It is not necessary to find the equation of the line. See Example 3.] 3 35. Slope - ; point 12, - 42 33. Slope 4; point 11, 22 34. Slope 2; point 1 - 2, 32 2 4 36. Slope ; point 1 - 3, 22 37. Slope - 2; point 1 - 2, - 32 38. Slope - 1; point 14, 12 3 In Problems 39–46, find an equation of the line L. 39. 40. y 2 (2, 1) L (–2, 1) 2 –1 (–1, 3) 2 x –2 –1 42. y 3 (–1, 1) x 2 L y –2 –2 2 L (2, 2) (1, 1) (0, 0) (0, 0) –2 41. L y –1 2 x –1 2 x SECTION 2.3 Lines y 43. y 44. 3 y 3 45. 3 (1, 2) (3, 3) 46. –1 y = 2x L –1 3 x L –1 y = 2x 3 x y = –x (–1, 1) 3 x –3 L L is perpendicular to y = 2x L is parallel to y = –x L is parallel to y = 2x y 3 (1, 2) L 179 1 x y = –x L is perpendicular to y = –x In Problems 47–72, find an equation for the line with the given properties. Express your answer using either the general form or the slope–intercept form of the equation of a line, whichever you prefer. 47. Slope = 3; containing the point 1 - 2, 32 48. Slope = 2; containing the point 14, - 32 2 49. Slope = - ; containing the point 11, - 12 3 50. Slope = 1 ; containing the point 13, 12 2 51. Containing the points 11, 32 and 1 - 1, 22 52. Containing the points 1 - 3, 42 and 12, 52 53. Slope = - 3; y@intercept = 3 54. Slope = - 2; y@intercept = - 2 55. x@intercept = 2; y@intercept = - 1 56. x@intercept = - 4; y@intercept = 4 57. Slope undefined; containing the point 12, 42 58. Slope undefined; containing the point 13, 82 59. Horizontal; containing the point 1 - 3, 22 60. Vertical; containing the point 14, - 52 61. Parallel to the line y = 2x; containing the point 1 - 1, 22 62. Parallel to the line y = - 3x; containing the point 1 - 1, 22 63. Parallel to the line 2x - y = - 2; containing the point 10, 02 64. Parallel to the line x - 2y = - 5; containing the point 10, 02 65. Parallel to the line x = 5; containing the point 14, 22 66. Parallel to the line y = 5; containing the point 14, 22 1 67. Perpendicular to the line y = x + 4; containing the point 2 11, - 22 68. Perpendicular to the line y = 2x - 3; containing the point 11, - 22 69. Perpendicular to the line 2x + y = 2; containing the point 1 - 3, 02 70. Perpendicular to the line x - 2y = - 5; containing the point 10, 42 71. Perpendicular to the line x = 8; containing the point 13, 42 72. Perpendicular to the line y = 8; containing the point 13, 42 In Problems 73–92, find the slope and y-intercept of each line. Graph the line. 1 73. y = 2x + 3 74. y = - 3x + 4 75. y = x - 1 2 1 79. x + 2y = 4 80. - x + 3y = 6 78. y = 2x + 2 85. x = - 4 84. x - y = 2 83. x + y = 1 88. x = 2 89. y - x = 0 90. x + y = 0 76. 1 x + y = 2 3 77. y = 1 x + 2 2 81. 2x - 3y = 6 82. 3x + 2y = 6 86. y = - 1 87. y = 5 91. 2y - 3x = 0 92. 3x + 2y = 0 In Problems 93–102, (a) find the intercepts of the graph of each equation and (b) graph the equation. 93. 2x + 3y = 6 94. 3x - 2y = 6 95. - 4x + 5y = 40 96. 6x - 4y = 24 97. 7x + 2y = 21 98. 5x + 3y = 18 99. 1 1 x + y = 1 2 3 100. x - 103. Find an equation of the x-axis. 2 y = 4 3 101. 0.2x - 0.5y = 1 102. - 0.3x + 0.4y = 1.2 104. Find an equation of the y-axis. In Problems 105–108, the equations of two lines are given. Determine whether the lines are parallel, perpendicular, or neither. 1 108. y = - 2x + 3 105. y = 2x - 3 107. y = 4x + 5 106. y = x - 3 2 y = 2x + 4 y = - 4x + 2 1 y = - 2x + 4 y = - x + 2 2 180 CHAPTER 2 Graphs In Problems 109–112, write an equation of each line. Express your answer using either the general form or the slope–intercept form of the equation of a line, whichever you prefer. 109. 110. 4 26 111. 2 2 6 23 24 112. 3 22 23 2 3 22 23 3 22 Applications and Extensions 113. Geometry Use slopes to show that the triangle whose vertices are 1 - 2, 52 , 11, 32 , and 1 - 1, 02 is a right triangle. 114. Geometry Use slopes to show that the quadrilateral whose vertices are 11, - 12 , 14, 12 , 12, 22 , and 15, 42 is a parallelogram. 115. Geometry Use slopes to show that the quadrilateral whose vertices are 1 - 1, 02 , 12, 32 , 11, - 22 , and 14, 12 is a rectangle. 116. Geometry Use slopes and the distance formula to show that the quadrilateral whose vertices are 10, 02 , 11, 32 , 14, 22 , and 13, - 12 is a square. 117. Truck Rentals A truck rental company rents a moving truck for one day by charging $39 plus $0.60 per mile. Write a linear equation that relates the cost C, in dollars, of renting the truck to the number x of miles driven. What is the cost of renting the truck if the truck is driven 110 miles? 230 miles? 118. Cost Equation The fixed costs of operating a business are the costs incurred regardless of the level of production. Fixed costs include rent, fixed salaries, and costs of leasing machinery. The variable costs of operating a business are the costs that change with the level of output. Variable costs include raw materials, hourly wages, and electricity. Suppose that a manufacturer of jeans has fixed daily costs of $500 and variable costs of $8 for each pair of jeans manufactured. Write a linear equation that relates the daily cost C, in dollars, of manufacturing the jeans to the number x of jeans manufactured. What is the cost of manufacturing 400 pairs of jeans? 740 pairs? 119. Cost of Driving a Car The annual fixed costs of owning a small sedan are $4462, assuming the car is completely paid for. The cost to drive the car is approximately $0.17 per mile. Write a linear equation that relates the cost C and the number x of miles driven annually. Source: AAA, April 2013 120. Wages of a Car Salesperson Dan receives $375 per week for selling new and used cars at a car dealership in Oak Lawn, Illinois. In addition, he receives 5% of the profit on any sales that he generates. Write a linear equation that represents Dan’s weekly salary S when he has sales that generate a profit of x dollars. 121. Electricity Rates in Illinois Commonwealth Edison Company supplies electricity to residential customers for a monthly customer charge of $15.37 plus 8.21 cents per kilowatt-hour for up to 800 kilowatt-hours (kW-hr). (a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, 0 … x … 800. (b) Graph this equation. (c) What is the monthly charge for using 200 kilowatthours? (d) What is the monthly charge for using 500 kilowatthours? (e) Interpret the slope of the line. Source: Commonwealth Edison Company, January 2014. 122. Electricity Rates in Florida Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $7.24 plus 9.07 cents per kilowatt-hour for up to 1000 kilowatt-hours. (a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, 0 … x … 1000. (b) Graph this equation. (c) What is the monthly charge for using 200 kilowatthours? (d) What is the monthly charge for using 500 kilowatthours? (e) Interpret the slope of the line. Source: Florida Power & Light Company, March 2014. 123. Measuring Temperature The relationship between Celsius (°C) and Fahrenheit (°F) degrees of measuring temperature is linear. Find a linear equation relating °C and °F if 0°C corresponds to 32°F and 100°C corresponds to 212°F. Use the equation to find the Celsius measure of 70°F. 124. Measuring Temperature The Kelvin (K) scale for measuring temperature is obtained by adding 273 to the Celsius temperature. SECTION 2.3 Lines (a) Write a linear equation relating K and °C. (b) Write a linear equation relating K and °F (see Problem 123). 125. Access Ramp A wooden access ramp is being built to reach a platform that sits 30 inches above the floor. The ramp drops 2 inches for every 25-inch run. y Platform 30 in. Ramp x (a) Write a linear equation that relates the height y of the ramp above the floor to the horizontal distance x from the platform. (b) Find and interpret the x-intercept of the graph of your equation. (c) Design requirements stipulate that the maximum run be 30 feet and that the maximum slope be a drop of 1 inch for each 12 inches of run. Will this ramp meet the requirements? Explain. (d) What slopes could be used to obtain the 30-inch rise and still meet design requirements? Source: www.adaptiveaccess.com/wood_ramps.php 126. Cigarette Use A report in the Child Trends DataBase indicated that in 2000, 20.6% of twelfth grade students reported daily use of cigarettes. In 2012, 9.3% of twelfth grade students reported daily use of cigarettes. (a) Write a linear equation that relates the percent y of twelfth grade students who smoke cigarettes daily to the number x of years after 2000. 181 (b) Find the intercepts of the graph of your equation. (c) Do these intercepts have meaningful interpretation? (d) Use your equation to predict the percent for the year 2025. Is this result reasonable? Source: www.childtrendsdatabank.org 127. Product Promotion A cereal company finds that the number of people who will buy one of its products in the first month that the product is introduced is linearly related to the amount of money it spends on advertising. If it spends $40,000 on advertising, then 100,000 boxes of cereal will be sold, and if it spends $60,000, then 200,000 boxes will be sold. (a) Write a linear equation that relates the amount A spent on advertising to the number x of boxes the company aims to sell. (b) How much expenditure on advertising is needed to sell 300,000 boxes of cereal? (c) Interpret the slope. 128. Show that the line containing the points 1a, b2 and 1b, a2, a ≠ b, is perpendicular to the line y = x. Also show that the midpoint of 1a, b2 and 1b, a2 lies on the line y = x. 129. The equation 2x - y = C defines a family of lines, one line for each value of C. On one set of coordinate axes, graph the members of the family when C = - 4, C = 0, and C = 2. Can you draw a conclusion from the graph about each member of the family? 130. Prove that if two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular. [Hint: Refer to Figure 47 and use the converse of the Pythagorean Theorem.] Explaining Concepts: Discussion and Writing 131. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 y (b) - 2x + 3y = 6 (c) 3x - 4y = - 12 (d) x - y = 1 (e) x - y = - 1 x (f) y = 3x - 5 (g) y = 2x + 3 (h) y = - 3x + 3 132. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 y (b) 2x - 3y = 6 (c) 3x + 4y = 12 (d) x - y = 1 x (e) x - y = - 1 (f) y = - 2x - 1 1 (g) y = - x + 10 2 (h) y = x + 4 133. The figure shows the graph of two parallel lines. Which of the following pairs of equations might have such a graph? (a) x - 2y = 3 y x + 2y = 7 (b) x + y = 2 x + y = -1 x (c) x - y = - 2 x - y = 1 (d) x - y = - 2 2x - 2y = - 4 (e) x + 2y = 2 x + 2y = - 1 134. The figure shows the graph of two perpendicular lines. Which of the following pairs of equations might have such a graph? (a) y - 2x = 2 y y + 2x = - 1 (b) y - 2x = 0 2y + x = 0 x (c) 2y - x = 2 2y + x = - 2 (d) y - 2x = 2 x + 2y = - 1 (e) 2x + y = - 2 2y + x = - 2 135. m is for Slope The accepted symbol used to denote the slope of a line is the letter m. Investigate the origin of this practice. Begin by consulting a French dictionary and looking up the French word monter. Write a brief essay on your findings. 182 CHAPTER 2 Graphs 136. Grade of a Road The term grade is used to describe the inclination of a road. How is this term related to the notion of slope of a line? Is a 4% grade very steep? Investigate the grades of some mountainous roads and determine their slopes. Write a brief essay on your findings. 138. Can the equation of every line be written in slope–intercept form? Why? 139. Does every line have exactly one x-intercept and one y-intercept? Are there any lines that have no intercepts? 140. What can you say about two lines that have equal slopes and equal y-intercepts? 141. What can you say about two lines with the same x-intercept and the same y-intercept? Assume that the x-intercept is not 0. Steep 7% Grade 142. If two distinct lines have the same slope but different x-intercepts, can they have the same y-intercept? 143. If two distinct lines have the same y-intercept but different slopes, can they have the same x-intercept? 144. Which form of the equation of a line do you prefer to use? Justify your position with an example that shows that your choice is better than another. Have reasons. 137. Carpentry Carpenters use the term pitch to describe the steepness of staircases and roofs. How is pitch related to slope? Investigate typical pitches used for stairs and for roofs. Write a brief essay on your findings. 145. What Went Wrong? A student is asked to find the slope of the line joining ( - 3, 2) and (1, - 4). He states that the slope 3 is . Is he correct? If not, what went wrong? 2 Retain Your Knowledge Problems 146–149 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 146. Simplify a x2y -3 b . Assume x ≠ 0 and y ≠ 0. Express the -2 x4y5 answer so that all exponents are positive. 147. The lengths of the legs of a right triangle are a = 8 and b = 15. Find the hypotenuse. 148. Solve the equation: (x - 3)2 + 25 = 49 149. Solve 2x - 5 + 7 6 10. Express the answer using set notation or interval notation. Graph the solution set. 2.4 Circles PREPARING FOR THIS SECTION Before getting started, review the following: r Completing the Square (Chapter R, Section R.5, p. 56) r Square Root Method (Section 1.2, pp. 94–95) Now Work the ‘Are You Prepared?’ problems on page 186. OBJECTIVES 1 Write the Standard Form of the Equation of a Circle (p. 182) 2 Graph a Circle (p. 183) 3 Work with the General Form of the Equation of a Circle (p. 184) 1 Write the Standard Form of the Equation of a Circle One advantage of a coordinate system is that it enables us to translate a geometric statement into an algebraic statement, and vice versa. Consider, for example, the following geometric statement that defines a circle. DEFINITION A circle is a set of points in the xy-plane that are a fixed distance r from a fixed point 1h, k2. The fixed distance r is called the radius, and the fixed point 1h, k2 is called the center of the circle. SECTION 2.4 Circles y 183 Figure 49 shows the graph of a circle. To find the equation, let 1x, y2 represent the coordinates of any point on a circle with radius r and center 1h, k2 . Then the distance between the points 1x, y2 and 1h, k2 must always equal r. That is, by the distance formula, (x, y) r 2 1x - h2 2 + 1y - k2 2 = r (h, k ) x or, equivalently, 1x - h2 2 + 1y - k2 2 = r 2 Figure 49 1x - h2 2 + 1y - k2 2 = r 2 DEFINITION The standard form of an equation of a circle with radius r and center 1h, k2 is 1x - h2 2 + 1y - k2 2 = r 2 THEOREM (1) The standard form of an equation of a circle of radius r with center at the origin 10, 02 is x2 + y 2 = r 2 DEFINITION If the radius r = 1, the circle whose center is at the origin is called the unit circle and has the equation x2 + y 2 = 1 See Figure 50. Notice that the graph of the unit circle is symmetric with respect to the x-axis, the y-axis, and the origin. y 1 1 Figure 50 Unit circle x2 + y2 = 1 EXAMPL E 1 Solution (0,0) 1 x 1 Writing the Standard Form of the Equation of a Circle Write the standard form of the equation of the circle with radius 5 and center 1 - 3, 62. Substitute the values r = 5, h = - 3, and k = 6 into equation (1). 1x - h2 2 + 1y - k2 2 = r 2 1x + 32 2 + 1y - 62 2 = 25 Now Work PROBLEM 9 r 2 Graph a Circle EXAMPL E 2 Solution Graphing a Circle Graph the equation: 1x + 32 2 + 1y - 22 2 = 16 Since the equation is in the form of equation (1), its graph is a circle. To graph the equation, compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle. 184 CHAPTER 2 Graphs (–3, 6) 1x + 32 2 + 1y - 22 2 = 16 y 1x - 1 - 32 2 2 + 1y - 22 2 = 42 6 4 (–7, 2) –10 (–3, 2) c (1, 2) 2 x –5 (–3, –2) Figure 51 1x + 32 2 + 1y - 22 2 = 16 EX AMPLE 3 Solution c c 1x - h2 2 + 1y - k2 2 = r 2 We see that h = - 3, k = 2, and r = 4. The circle has center 1 - 3, 22 and a radius of 4 units. To graph this circle, first plot the center 1 - 3, 22 . Since the radius is 4, we can locate four points on the circle by plotting points 4 units to the left, to the right, up, and down from the center. These four points can then be used as guides to obtain the graph. See Figure 51. r Now Work PROBLEMS 25(a) AND (b) Finding the Intercepts of a Circle For the circle 1x + 32 2 + 1y - 22 2 = 16, find the intercepts, if any, of its graph. This is the equation discussed and graphed in Example 2. To find the x-intercepts, if any, let y = 0. Then 1x + 32 2 + 1y - 22 2 = 16 In Words The symbol { is read “plus or minus.” It means to add and subtract the quantity following the { symbol. For example, 5 { 2 means “5 - 2 = 3 or 5 + 2 = 7.” 1x + 32 2 + 10 - 22 2 = 16 y = 0 1x + 32 + 4 = 16 2 Simplify. 1x + 32 = 12 2 Simplify. x + 3 = { 212 Apply the Square Root Method. x = - 3 { 223 Solve for x. The x-intercepts are - 3 - 223 ≈ - 6.46 and - 3 + 223 ≈ 0.46. To find the y-intercepts, if any, let x = 0. Then 1x + 32 2 + 1y - 22 2 = 16 10 + 32 2 + 1y - 22 2 = 16 9 + 1y - 22 2 = 16 1y - 22 2 = 7 y - 2 = { 27 y = 2 { 27 The y-intercepts are 2 - 27 ≈ - 0.65 and 2 + 27 ≈ 4.65. Look back at Figure 51 to verify the approximate locations of the intercepts. Now Work PROBLEM 25(C) r 3 Work with the General Form of the Equation of a Circle If we eliminate the parentheses from the standard form of the equation of the circle given in Example 2, we get 1x + 32 2 + 1y - 22 2 = 16 x2 + 6x + 9 + y2 - 4y + 4 = 16 which simplifies to x2 + y2 + 6x - 4y - 3 = 0 (2) It can be shown that any equation of the form x2 + y2 + ax + by + c = 0 has a graph that is a circle, is a point, or has no graph at all. For example, the graph of the equation x2 + y2 = 0 is the single point 10, 02. The equation x2 + y2 + 5 = 0, or x2 + y2 = - 5, has no graph, because sums of squares of real numbers are never negative. SECTION 2.4 Circles DEFINITION 185 When its graph is a circle, the equation x2 + y2 + ax + by + c = 0 is the general form of the equation of a circle. Now Work PROBLEM 15 If an equation of a circle is in general form, we use the method of completing the square to put the equation in standard form so that we can identify its center and radius. EXAMPL E 4 Graphing a Circle Whose Equation Is in General Form Graph the equation: x2 + y2 + 4x - 6y + 12 = 0 Solution Group the terms involving x, group the terms involving y, and put the constant on the right side of the equation. The result is 1x2 + 4x2 + 1y2 - 6y2 = - 12 y (–2, 4) 4 1 (–3, 3) Next, complete the square of each expression in parentheses. Remember that any number added on the left side of the equation must also be added on the right. 1x2 + 4x + 42 + 1y2 - 6y + 92 = - 12 + 4 + 9 (–1, 3) c 6 4 2 a b = 4 2 (–2, 3) (–2, 2) c 6 a -6 2 b = 9 2 1x + 22 2 + 1y - 32 2 = 1 Factor. 1 x –3 Figure 52 1x + 22 2 + 1y - 32 2 = 1 This equation is the standard form of the equation of a circle with radius 1 and center 1 - 2, 32 . To graph the equation, use the center 1 - 2, 32 and the radius 1. See Figure 52. Now Work EXAMPL E 5 PROBLEM 29 r Using a Graphing Utility to Graph a Circle Graph the equation: x2 + y2 = 4 Solution Y1 5 √42x 2 This is the equation of a circle with center at the origin and radius 2. To graph this equation, solve for y. x2 + y 2 = 4 y 2 = 4 - x2 2.5 y = { 24 - x 24 4 22.5 Y2 5 2√42x 2 Figure 53 x + y = 4 2 2 Subtract x 2 from each side. 2 Apply the Square Root Method to solve for y. There are two equations to graph: first graph Y1 = 24 - x2 and then graph Y2 = - 24 - x2 on the same square screen. (Your circle will appear oval if you do not use a square screen.*) See Figure 53. Overview The discussion in Sections 2.3 and 2.4 about lines and circles dealt with two main types of problems that can be generalized as follows: 1. Given an equation, classify it and graph it. 2. Given a graph, or information about a graph, find its equation. This text deals with both types of problems. We shall study various equations, classify them, and graph them. The second type of problem is usually more difficult to solve than the first. *The square screen ratio for the TI-84 Plus C calculator is 8:5. 186 CHAPTER 2 Graphs 2.4 Assess Your Understanding Are You Prepared? Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. To complete the square of x2 + 10x, you would (add/subtract) the number . (p. 56) 2. Use the Square Root Method to solve the equation 1x - 22 2 = 9. (pp. 94–95) Concepts and Vocabulary 3. True or False Every equation of the form 7. Choose the equation of a circle with radius 6 and center (3, –5). (a) 1x - 32 2 + 1y + 52 2 = 6 (b) 1x + 32 2 + 1y - 52 2 = 36 (c) 1x + 32 2 + 1y - 52 2 = 6 (d) 1x - 32 2 + 1y + 52 2 = 36 x + y + ax + by + c = 0 2 2 has a circle as its graph. 4. For a circle, the any point on the circle. is the distance from the center to 8. The equation of a circle can be changed from general form to standard from by doing which of the following? (a) completing the squares (b) solving for x (c) solving for y (d) squaring both sides 5. True or False The radius of the circle x2 + y2 = 9 is 3. 6. True or False The center of the circle 1x + 32 2 + 1y - 22 2 = 13 is (3, - 2). Skill Building In Problems 9–12, find the center and radius of each circle. Write the standard form of the equation. 10. 9. y 11. y y (4, 2) 12. y (2, 3) (1, 2) (0, 1) (2, 1) (0, 1) (1, 2) x (1, 0) x x x In Problems 13–22, write the standard form of the equation and the general form of the equation of each circle of radius r and center 1h, k2. Graph each circle. 13. r = 2; 17. r = 5; 21. r = 1 ; 2 1h, k2 = 10, 02 1h, k2 = 14, - 32 14. r = 3; 18. r = 4; 1h, k2 = 10, 02 1h, k2 = 12, - 32 1 1h, k2 = a , 0b 2 15. r = 2; 19. r = 4; 22. r = 1 ; 2 1h, k2 = 10, 22 1h, k2 = 1 - 2, 12 16. r = 3; 20. r = 7; 1h, k2 = 11, 02 1h, k2 = 1 - 5, - 22 1 1h, k2 = a0, - b 2 In Problems 23–36, (a) find the center 1h, k2 and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any. 23. x2 + y2 = 4 24. x2 + 1y - 12 2 = 1 25. 21x - 32 2 + 2y2 = 8 26. 31x + 12 2 + 31y - 12 2 = 6 27. x2 + y2 - 2x - 4y - 4 = 0 28. x2 + y2 + 4x + 2y - 20 = 0 29. x2 + y2 + 4x - 4y - 1 = 0 1 32. x2 + y2 + x + y - = 0 2 35. 2x2 + 8x + 2y2 = 0 30. x2 + y2 - 6x + 2y + 9 = 0 31. x2 + y2 - x + 2y + 1 = 0 33. 2x2 + 2y2 - 12x + 8y - 24 = 0 34. 2x2 + 2y2 + 8x + 7 = 0 36. 3x2 + 3y2 - 12y = 0 In Problems 37–44, find the standard form of the equation of each circle. 37. Center at the origin and containing the point 1 - 2, 32 38. Center 11, 02 and containing the point 1 - 3, 22 39. Center 12, 32 and tangent to the x-axis 40. Center 1 - 3, 12 and tangent to the y-axis 43. Center 1 - 1, 32 and tangent to the line y = 2 44. Center 14, - 22 and tangent to the line x = 1 41. With endpoints of a diameter at 11, 42 and 1 - 3, 22 42. With endpoints of a diameter at 14, 32 and 10, 12 SECTION 2.4 Circles 187 In Problems 45–48, match each graph with the correct equation. (a) 1x - 32 2 + 1y + 32 2 = 9 (b) 1x + 12 2 + 1y - 22 2 = 4 (c) 1x - 12 2 + 1y + 22 2 = 4 (d) 1x + 32 2 + 1y - 32 2 = 9 45. 46. 47. 4 6 26.4 6.4 29.6 48. 4 6 9.6 26.4 24 26 6.4 29.6 9.6 24 26 Applications and Extensions 53. Weather Satellites Earth is represented on a map of a portion of the solar system so that its surface is the circle with equation x2 + y2 + 2x + 4y - 4091 = 0. A weather satellite circles 0.6 unit above Earth with the center of its circular orbit at the center of Earth. Find the equation for the orbit of the satellite on this map. 49. Find the area of the square in the figure. y x2 y2 9 x r 50. Find the area of the blue shaded region in the figure, assuming the quadrilateral inside the circle is a square. y x 2 y 2 36 54. The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. See the figure. x y r 51. Ferris Wheel The original Ferris wheel was built in 1893 by Pittsburgh, Pennsylvania bridge builder George W. Ferris. The Ferris wheel was originally built for the 1893 World’s Fair in Chicago, but it was also later reconstructed for the 1904 World’s Fair in St. Louis. It had a maximum height of 264 feet and a wheel diameter of 250 feet. Find an equation for the wheel if the center of the wheel is on the y-axis. Source: guinnessworldrecords.com 52. Ferris Wheel Opening in 2014 in Las Vegas, The High Roller observation wheel has a maximum height of 550 feet and a diameter of 520 feet, with one full rotation taking approximately 30 minutes. Find an equation for the wheel if the center of the wheel is on the y-axis. Source: Las Vegas Review Journal x If the equation of the circle is x2 + y2 = r 2 and the equation of the tangent line is y = mx + b, show that: (a) r 2 11 + m2 2 = b2 [Hint: The quadratic equation x2 + 1mx + b2 2 = r 2 has exactly one solution.] - r 2m r 2 (b) The point of tangency is ¢ , ≤. b b (c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency. 55. The Greek Method The Greek method for finding the equation of the tangent line to a circle uses the fact that at any point on a circle, the lines containing the center and the tangent line are perpendicular (see Problem 54). Use this method to find an equation of the tangent line to the circle x2 + y2 = 9 at the point 11, 2122 . 56. Use the Greek method described in Problem 55 to find an equation of the tangent line to the circle x2 + y2 - 4x + 6y + 4 = 0 at the point 13, 212 - 32 . 188 CHAPTER 2 Graphs 57. Refer to Problem 54. The line x - 2y + 4 = 0 is tangent to a circle at 10, 22 . The line y = 2x - 7 is tangent to the same circle at 13, - 12 . Find the center of the circle. 59. If a circle of radius 2 is made to roll along the x-axis, what is an equation for the path of the center of the circle? 60. If the circumference of a circle is 6p, what is its radius? 58. Find an equation of the line containing the centers of the two circles x2 + y2 - 4x + 6y + 4 = 0 and x2 + y2 + 6x + 4y + 9 = 0 Explaining Concepts: Discussion and Writing 61. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 1x - 22 2 + 1y + 32 2 = 13 y (b) 1x - 22 2 + 1y - 22 2 = 8 (c) 1x - 22 2 + 1y - 32 2 = 13 (d) 1x + 22 2 + 1y - 22 2 = 8 (e) x2 + y2 - 4x - 9y = 0 x (f) x2 + y2 + 4x - 2y = 0 (g) x2 + y2 - 9x - 4y = 0 (h) x2 + y2 - 4x - 4y = 4 62. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 1x - 22 2 + y2 = 3 y (b) 1x + 22 2 + y2 = 3 (c) x2 + 1y - 22 2 = 3 (d) 1x + 22 2 + y2 = 4 (e) x2 + y2 + 10x + 16 = 0 x (f) x2 + y2 + 10x - 2y = 1 (g) x2 + y2 + 9x + 10 = 0 (h) x2 + y2 - 9x - 10 = 0 63. Explain how the center and radius of a circle can be used to graph the circle. 64. What Went Wrong? A student stated that the center and radius of the graph whose equation is (x + 3)2 + (y - 2)2 = 16 are (3, - 2) and 4, respectively. Why is this incorrect? Retain Your Knowledge Problems 65–68 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 65. Find the area and circumference of a circle of radius 13 cm. 2 66. Multiply (3x - 2)(x - 2x + 3). Express the answer as a polynomial in standard form. 68. Aaron can load a delivery van in 22 minutes. Elizabeth can load the same van in 28 minutes. How long would it take them to load the van if they worked together? 67. Solve the equation: 22x2 + 3x - 1 = x + 1 ‘Are You Prepared?’ Answers 1. add; 25 2. 5 - 1, 56 2.5 Variation OBJECTIVES 1 Construct a Model Using Direct Variation (p. 189) 2 Construct a Model Using Inverse Variation (p. 189) 3 Construct a Model Using Joint Variation or Combined Variation (p. 190) When a mathematical model is developed for a real-world problem, it often involves relationships between quantities that are expressed in terms of proportionality: Force is proportional to acceleration. When an ideal gas is held at a constant temperature, pressure and volume are inversely proportional. The force of attraction between two heavenly bodies is inversely proportional to the square of the distance between them. Revenue is directly proportional to sales. SECTION 2.5 Variation 189 Each of the preceding statements illustrates the idea of variation, or how one quantity varies in relation to another quantity. Quantities may vary directly, inversely, or jointly. 1 Construct a Model Using Direct Variation DEFINITION Let x and y denote two quantities. Then y varies directly with x, or y is directly proportional to x, if there is a nonzero number k such that y = kx The number k is called the constant of proportionality. y The graph in Figure 54 illustrates the relationship between y and x if y varies directly with x and k 7 0, x Ú 0. Note that the constant of proportionality is, in fact, the slope of the line. If two quantities vary directly, then knowing the value of each quantity in one instance enables us to write a formula that is true in all cases. x Figure 54 y = k x ; k 7 0, x Ú 0 Mortgage Payments EXAMPL E 1 The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6.65 for every $1000 borrowed, find a formula that relates the monthly payment p to the amount borrowed B for a mortgage with these terms. Then find the monthly payment p when the amount borrowed B is $120,000. Solution Because p varies directly with B, we know that p = kB p for some constant k. Because p = 6.65 when B = 1000, it follows that Monthly Payment 800 6.65 = k 110002 (120 000, 798) 600 k = 0.00665 Since p = kB, 400 p = 0.00665B 200 0 Solve for k. 40 80 120 160 Amount Borrowed (000's) B In particular, when B = $120,000, p = 0.006651$120,0002 = $798 Figure 55 illustrates the relationship between the monthly payment p and the amount borrowed B. Figure 55 Now Work PROBLEMS 5 AND 23 r 2 Construct a Model Using Inverse Variation y DEFINITION Let x and y denote two quantities. Then y varies inversely with x, or y is inversely proportional to x, if there is a nonzero constant k such that y = k x x Figure 56 y = k ; k 7 0, x 7 0 x The graph in Figure 56 illustrates the relationship between y and x if y varies inversely with x and k 7 0, x 7 0. 190 CHAPTER 2 Graphs EX AMPLE 2 Maximum Weight That Can Be Supported by a Piece of Pine See Figure 57. The maximum weight W that can be safely supported by a 2-inch by 4-inch piece of pine varies inversely with its length l. Experiments indicate that the maximum weight that a 10-foot-long 2-by-4 piece of pine can support is 500 pounds. Write a general formula relating the maximum weight W (in pounds) to length l (in feet). Find the maximum weight W that can be safely supported by a length of 25 feet. Solution Because W varies inversely with l, we know that W = k l for some constant k. Because W = 500 when l = 10, we have k 10 500 = k = 5000 Since W = k , l W = Figure 57 5000 l In particular, the maximum weight W that can be safely supported by a piece of pine 25 feet in length is W 600 5000 = 200 pounds 25 W = (10, 500) 500 400 Figure 58 illustrates the relationship between the weight W and the length l. 300 Now Work 200 (25, 200) PROBLEM r 33 100 0 5 10 Figure 58 W = 15 20 25 l 5000 l 3 Construct a Model Using Joint Variation or Combined Variation When a variable quantity Q is proportional to two or more other variables, we say that Q varies jointly with these quantities. Finally, combinations of direct and/or inverse variation may occur. This is usually referred to as combined variation. EX AMPLE 3 Loss of Heat through a Wall The loss of heat through a wall varies jointly with the area of the wall and the difference between the inside and outside temperatures and varies inversely with the thickness of the wall. Write an equation that relates these quantities. Solution Begin by assigning symbols to represent the quantities: L = Heat loss T = Temperature difference A = Area of wall d = Thickness of wall Then L = k where k is the constant of proportionality. AT d r SECTION 2.5 Variation 191 In direct or inverse variation, the quantities that vary may be raised to powers. For example, in the early seventeenth century, Johannes Kepler (1571–1630) discovered that the square of the period of revolution T of a planet around the Sun varies directly with the cube of its mean distance a from the Sun. That is, T 2 = ka3, where k is the constant of proportionality. EXAMPL E 4 Force of the Wind on a Window The force F of the wind on a flat surface positioned at a right angle to the direction of the wind varies jointly with the area A of the surface and the square of the speed v of the wind. A wind of 30 miles per hour blowing on a window measuring 4 feet by 5 feet has a force of 150 pounds. See Figure 59. What force does a wind of 50 miles per hour exert on a window measuring 3 feet by 4 feet? Solution Since F varies jointly with A and v2, we have F = kAv2 where k is the constant of proportionality. We are told that F = 150 when A = 4 # 5 = 20 and v = 30. Then 150 = k 1202 19002 F = kAv 2, F = 150, A = 20, v = 30 Wind 1 120 k = Figure 59 Since F = kAv2, F = 1 Av2 120 For a wind of 50 miles per hour blowing on a window whose area is A = 3 # 4 = 12 square feet, the force F is F = Now Work 1 1122 125002 = 250 pounds 120 PROBLEM r 41 2.5 Assess Your Understanding Concepts and Vocabulary 1. If x and y are two quantities, then y is directly proportional to x if there is a nonzero number k such that . k 2. True or False If y varies directly with x, then y = , where x k is a constant. 3. Which equation represents a joint variation model? (a) y = 5x (b) y = 5xzw 5 5xz (c) y = (d) y = x w kx 4. Choose the best description for the model y = , if k is a z nonzero constant. (a) y varies jointly with x and z. (b) y is inversely proportional to x and z. (c) y varies directly with x and inversely with z. (d) y is directly proportion to z and inversely proportional to x. Skill Building In Problems 5–16, write a general formula to describe each variation. 5. y varies directly with x; y = 2 when x = 10 6. v varies directly with t; v = 16 when t = 2 7. A varies directly with x2; A = 4p when x = 2 8. V varies directly with x3; V = 36p when x = 3 9. F varies inversely with d 2; F = 10 when d = 5 10. y varies inversely with 1x; y = 4 when x = 9 11. z varies directly with the sum of the squares of x and y; z = 5 when x = 3 and y = 4 12. T varies jointly with the cube root of x and the square of d; T = 18 when x = 8 and d = 3 192 CHAPTER 2 Graphs 13. M varies directly with the square of d and inversely with the square root of x; M = 24 when x = 9 and d = 4 14. z varies directly with the sum of the cube of x and the square of y; z = 1 when x = 2 and y = 3 15. The square of T varies directly with the cube of a and inversely with the square of d; T = 2 when a = 2 and d = 4 16. The cube of z varies directly with the sum of the squares of x and y; z = 2 when x = 9 and y = 4 Applications and Extensions In Problems 17–22, write an equation that relates the quantities. 17. Geometry The volume V of a sphere varies directly with the 4p cube of its radius r. The constant of proportionality is . 3 18. Geometry The square of the length of the hypotenuse c of a right triangle varies jointly with the sum of the squares of the lengths of its legs a and b. The constant of proportionality is 1. 19. Geometry The area A of a triangle varies jointly with the product of the lengths of the base b and the height h. The 1 constant of proportionality is . 2 20. Geometry The perimeter p of a rectangle varies jointly with the sum of the lengths of its sides l and w. The constant of proportionality is 2. 21. Physics: Newton’s Law The force F (in newtons) of attraction between two bodies varies jointly with their masses m and M (in kilograms) and inversely with the square of the distance d (in meters) between them. The constant of proportionality is G = 6.67 * 10-11. 22. Physics: Simple Pendulum The period of a pendulum is the time required for one oscillation; the pendulum is usually referred to as simple when the angle made to the vertical is less than 5°. The period T of a simple pendulum (in seconds) varies directly with the square root of its length l (in feet). 2p The constant of proportionality is . 232 23. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6.49 for every $1000 borrowed, find a linear equation that relates the monthly payment p to the amount borrowed B for a mortgage with these terms. Then find the monthly payment p when the amount borrowed B is $145,000. 24. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 15-year mortgage is $8.99 for every $1000 borrowed, find a linear equation that relates the monthly payment p to the amount borrowed B for a mortgage with these terms. Then find the monthly payment p when the amount borrowed B is $175,000. 25. Physics: Falling Objects The distance s that an object falls is directly proportional to the square of the time t of the fall. If an object falls 16 feet in 1 second, how far will it fall in 3 seconds? How long will it take an object to fall 64 feet? 26. Physics: Falling Objects The velocity v of a falling object is directly proportional to the time t of the fall. If, after 2 seconds, the velocity of the object is 64 feet per second, what will its velocity be after 3 seconds? 27. Physics: Stretching a Spring The elongation E of a spring balance varies directly with the applied weight W (see the figure). If E = 3 when W = 20, find E when W = 15. E W 28. Physics: Vibrating String The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 48 inches long and vibrates 256 times per second, what is the length of a string that vibrates 576 times per second? 29. Revenue Equation At the corner Shell station, the revenue R varies directly with the number g of gallons of gasoline sold. If the revenue is $47.40 when the number of gallons sold is 12, find a linear equation that relates revenue R to the number g of gallons of gasoline sold. Then find the revenue R when the number of gallons of gasoline sold is 10.5. 30. Cost Equation The cost C of roasted almonds varies directly with the number A of pounds of almonds purchased. If the cost is $23.75 when the number of pounds of roasted almonds purchased is 5, find a linear equation that relates the cost C to the number A of pounds of almonds purchased. Then find the cost C when the number of pounds of almonds purchased is 3.5. 31. Demand Suppose that the demand D for candy at the movie theater is inversely related to the price p. (a) When the price of candy is $2.75 per bag, the theater sells 156 bags of candy. Express the demand for candy in terms of its price. (b) Determine the number of bags of candy that will be sold if the price is raised to $3 a bag. 32. Driving to School The time t that it takes to get to school varies inversely with your average speed s. (a) Suppose that it takes you 40 minutes to get to school when your average speed is 30 miles per hour. Express the driving time to school in terms of average speed. (b) Suppose that your average speed to school is 40 miles per hour. How long will it take you to get to school? 33. Pressure The volume of a gas V held at a constant temperature in a closed container varies inversely with its pressure P. If the volume of a gas is 600 cubic centimeters 1cm3 2 when the pressure is 150 millimeters of mercury (mm Hg), find the volume when the pressure is 200 mm Hg. SECTION 2.5 Variation 193 34. Resistance The current i in a circuit is inversely proportional to its resistance Z measured in ohms. Suppose that when the current in a circuit is 30 amperes, the resistance is 8 ohms. Find the current in the same circuit when the resistance is 10 ohms. surface and the square of the velocity of the wind. If the force on an area of 20 square feet is 11 pounds when the wind velocity is 22 miles per hour, find the force on a surface area of 47.125 square feet when the wind velocity is 36.5 miles per hour. 35. Weight The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of Earth. If Maria weighs 125 pounds when she is on the surface of Earth (3960 miles from the center), determine Maria’s weight when she is at the top of Mount McKinley (3.8 miles from the surface of Earth). 41. Horsepower The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute, rpm) and the cube of its diameter. If a shaft of a certain material 2 inches in diameter can transmit 36 hp at 75 rpm, what diameter must the shaft have in order to transmit 45 hp at 125 rpm? 36. Weight of a Body The weight of a body above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a certain body weighs 55 pounds when it is 3960 miles from the center of Earth, how much will it weigh when it is 3965 miles from the center? 42. Chemistry: Gas Laws The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. Write an equation relating V, T, and P using k as the constant of proportionality. If a cylinder contains oxygen at a temperature of 300 K and a pressure of 15 atmospheres in a volume of 100 liters, what is the constant of proportionality k? If a piston is lowered into the cylinder, decreasing the volume occupied by the gas to 80 liters and raising the temperature to 310 K, what is the gas pressure? 37. Geometry The volume V of a right circular cylinder varies jointly with the square of its radius r and its height h. The constant of proportionality is p. See the figure. Write an equation for V. r h 38. Geometry The volume V of a right circular cone varies jointly with the square of its radius r and its height h. The p constant of proportionality is . See the figure. Write an 3 equation for V. h r 39. Intensity of Light The intensity I of light (measured in foot-candles) varies inversely with the square of the distance from the bulb. Suppose that the intensity of a 100-watt light bulb at a distance of 2 meters is 0.075 foot-candle. Determine the intensity of the bulb at a distance of 5 meters. 40. Force of the Wind on a Window The force exerted by the wind on a plane surface varies jointly with the area of the 43. Physics: Kinetic Energy The kinetic energy K of a moving object varies jointly with its mass m and the square of its velocity v. If an object weighing 25 kilograms and moving with a velocity of 10 meters per second has a kinetic energy of 1250 joules, find its kinetic energy when the velocity is 15 meters per second. 44. Electrical Resistance of a Wire The electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire. If a wire 432 feet long and 4 millimeters in diameter has a resistance of 1.24 ohms, find the length of a wire of the same material whose resistance is 1.44 ohms and whose diameter is 3 millimeters. 45. Measuring the Stress of Materials The stress in the material of a pipe subject to internal pressure varies jointly with the internal pressure and the internal diameter of the pipe and inversely with the thickness of the pipe. The stress is 100 pounds per square inch when the diameter is 5 inches, the thickness is 0.75 inch, and the internal pressure is 25 pounds per square inch. Find the stress when the internal pressure is 40 pounds per square inch if the diameter is 8 inches and the thickness is 0.50 inch. 46. Safe Load for a Beam The maximum safe load for a horizontal rectangular beam varies jointly with the width of the beam and the square of the thickness of the beam and inversely with its length. If an 8-foot beam will support up to 750 pounds when the beam is 4 inches wide and 2 inches thick, what is the maximum safe load in a similar beam 10 feet long, 6 inches wide, and 2 inches thick? Explaining Concepts: Discussion and Writing 47. In the early seventeenth century, Johannes Kepler discovered that the square of the period T of the revolution of a planet around the Sun varies directly with the cube of its mean distance a from the Sun. Go to the library and research this law and Kepler’s other two laws. Write a brief paper about these laws and Kepler’s place in history. 48. Using a situation that has not been discussed in the text, write a real-world problem that you think involves two variables that vary directly. Exchange your problem with another student’s to solve and critique. 49. Using a situation that has not been discussed in the text, write a real-world problem that you think involves two variables that vary inversely. Exchange your problem with another student’s to solve and critique. 50. Using a situation that has not been discussed in the text, write a real-world problem that you think involves three variables that vary jointly. Exchange your problem with another student’s to solve and critique. 194 CHAPTER 2 Graphs Retain Your Knowledge Problems 51–54 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 4 3>2 51. Factor 3x3 + 25x2 - 12x - 100 completely. 53. Simplify: a b 25 3 5 x - 2 . 54. Rationalize the denominator of 52. Add and simplify the result. + 2 x + 3 x + 7x + 12 27 - 2 Chapter Review Things to Know Formulas Distance formula (p. 151) d = 21x2 - x1 2 2 + 1y2 - y1 2 2 Midpoint formula (p. 154) 1x, y2 = a Slope (p. 167) Parallel lines (p. 175) x1 + x2 y1 + y2 , b 2 2 y2 - y1 if x1 ≠ x2; undefined if x1 = x2 m = x2 - x1 Equal slopes 1m1 = m2 2 and different y-intercepts 1b1 ≠ b2 2 Perpendicular lines (p. 176) Product of slopes is - 1 1m1 # m2 = - 12 Direct variation (p. 189) y = kx Inverse variation (p. 189) y = k x Equations of Lines and Circles Vertical line (p. 171) x = a; a is the x-intercept Horizontal line (p. 172) y = b; b is the y-intercept Point–slope form of the equation of a line (p. 171) y - y1 = m1x - x1 2; m is the slope of the line, 1x1 , y1 2 is a point on the line Slope–intercept form of the equation of a line (p. 172) y = mx + b; m is the slope of the line, b is the y-intercept General form of the equation of a line (p. 174) Ax + By = C; A, B not both 0 Standard form of the equation of a circle (p. 183) 1x - h2 2 + 1y - k2 2 = r 2; r is the radius of the circle, 1h, k2 is the center of the circle Equation of the unit circle (p. 183) x2 + y 2 = 1 General form of the equation of a circle (p. 185) x2 + y2 + ax + by + c = 0, with restrictions on a, b, and c Objectives Section You should be able to … Examples Review Exercises 2.1 Use the distance formula (p. 151) Use the midpoint formula (p. 153) 1–3 4 1(a)–3(a), 29, 30(a), 31 1(b)–3(b), 31 Graph equations by plotting points (p. 157) Find intercepts from a graph (p. 159) 1–3 4 4 5 1 2 2.2 1 2 2.3 3 Find intercepts from an equation (p. 160) 5 6–10 4 6–9 6–10 5 Test an equation for symmetry with respect to the x-axis, the y-axis, and the origin (p. 160) Know how to graph key equations (p. 163) 10–12 26, 27 1 Calculate and interpret the slope of a line (p. 167) 1, 2 2 Graph lines given a point and the slope (p. 170) 3 1(c)–3(c), 1(d)–3(d), 32 28 3 Find the equation of a vertical line (p. 170) 4 17 Chapter Review Section You should be able to . . . Examples Review Exercises Use the point–slope form of a line; identify horizontal lines (p. 171) 5, 6 16 5 Find the equation of a line given two points (p. 172) 7 18, 19 6 Write the equation of a line in slope–intercept form (p. 172) 8 16, 18–21 7 Identify the slope and y-intercept of a line from its equation (p. 173) 8 22, 23 8 Graph lines written in general form using intercepts (p. 174) 9 24, 25 9 Find equations of parallel lines (p. 175) 10, 11 20 Find equations of perpendicular lines (p. 176) 12, 13 21, 30(b) 2 Write the standard form of the equation of a circle (p. 182) Graph a circle (p. 183) 1 2, 3, 5 11, 12, 31 13–15 3 Work with the general form of the equation of a circle (p. 184) 4 14, 15 1 2 Construct a model using direct variation (p. 189) Construct a model using inverse variation (p. 189) 1 2 33 34 3 Construct a model using joint or combined variation (p. 190) 3, 4 35 4 10 2.4 2.5 1 195 Review Exercises In Problems 1–3, find the following for each pair of points: (a) The distance between the points (b) The midpoint of the line segment connecting the points (c) The slope of the line containing the points (d) Interpret the slope found in part (c) 5. List the intercepts of the graph below. y 1. 10, 02; 14, 22 2 2. 11, - 12; 1 - 2, 32 4 3. 14, - 42; 14, 82 4 x 2 4. Graph y = x2 + 4 by plotting points. In Problems 6–10, list the intercepts and test for symmetry with respect to the x-axis, the y-axis, and the origin. 7. x2 + 4y2 = 16 6. 2x = 3y2 8. y = x4 + 2x2 + 1 10. x2 + x + y2 + 2y = 0 9. y = x3 - x In Problems 11 and 12, find the standard form of the equation of the circle whose center and radius are given. 11. 1h, k2 = 1 - 2, 32; r = 4 12. 1h, k2 = 1 - 1, - 22; r = 1 In Problems 13–15, find the center and radius of each circle. Graph each circle. Find the intercepts, if any, of each circle. 13. x2 + 1y - 12 2 = 4 14. x2 + y2 - 2x + 4y - 4 = 0 15. 3x2 + 3y2 - 6x + 12y = 0 In Problems 16–21, find an equation of the line having the given characteristics. Express your answer using either the general form or the slope–intercept form of the equation of a line, whichever you prefer. 16. Slope = - 2; containing the point 13, - 12 18. y@intercept = - 2; containing the point 15, - 32 20. Parallel to the line 2x - 3y = - 4; 21. Perpendicular to the line x + y = 2; 17. Vertical; containing the point 1 - 3, 42 19. Containing the points 13, - 42 and 12, 12 containing the point 1 - 5, 32 containing the point 14, - 32 In Problems 22 and 23, find the slope and y-intercept of each line. Graph the line, labeling any intercepts. 1 1 1 22. 4x - 5y = - 20 23. x - y = 2 3 6 In Problems 24 and 25, find the intercepts and graph each line. 1 1 24. 2x - 3y = 12 25. x + y = 2 2 3 196 CHAPTER 2 Graphs 26. Sketch a graph of y = x3. 27. Sketch a graph of y = 2x. 2 28. Graph the line with slope containing the point 11, 22 . 3 29. Show that the points A = 13, 42, B = 11, 12 , and C = 1 - 2, 32 are the vertices of an isosceles triangle. 30. Show that the points A = 1 - 2, 02, B = 1 - 4, 42 , and C = 18, 52 are the vertices of a right triangle in two ways: (a) By using the converse of the Pythagorean Theorem (b) By using the slopes of the lines joining the vertices 31. The endpoints of the diameter of a circle are 1 - 3, 22 and 15, - 62 . Find the center and radius of the circle. Write the standard equation of this circle. 32. Show that the points A = 12, 52, B = 16, 12 , C = 18, - 12 lie on a line by using slopes. 34. Weight of a Body The weight of a body varies inversely with the square of its distance from the center of Earth. Assuming that the radius of Earth is 3960 miles, how much would a man weigh at an altitude of 1 mile above Earth’s surface if he weighs 200 pounds on Earth’s surface? 35. Heat Loss The amount of heat transferred per hour through a glass window varies jointly with the surface area of the window and the difference in temperature between the areas separated by the glass. A window with a surface area of 7.5 square feet loses 135 Btu per hour when the temperature difference is 40°F. How much heat is lost per hour for a similar window with a surface are of 12 square feet when the temperature difference is 35°F? and 33. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $854.00 when $130,000 is borrowed, find an equation that relates the monthly payment p to the amount borrowed B for a mortgage with these terms. Then find the monthly payment p when the amount borrowed B is $165,000. The Chapter Test Prep Videos are step-by-step solutions available in , or on this text’s Channel. Flip back to the Resources for Success page for a link to this text’s YouTube channel. Chapter Test In Problems 1–3, use P1 = 1 - 1, 32 and P2 = 15, - 12 . 1. Find the distance from P1 to P2. 2. Find the midpoint of the line segment joining P1 and P2. 3. (a) Find the slope of the line containing P1 and P2. (b) Interpret this slope. 4. Graph y = x2 - 9 by plotting points. 5. Sketch the graph of y2 = x. 6. List the intercepts and test for symmetry: x2 + y = 9 7. Write the slope–intercept form of the line with slope - 2 containing the point 13, - 42 . Graph the line. 8. Write the general form of the circle with center 14, - 32 and radius 5. 9. Find the center and radius of the circle x2 + y2 + 4x - 2y - 4 = 0. Graph this circle. 10. For the line 2x + 3y = 6, find a line parallel to it containing the point 11, - 12 . Also find a line perpendicular to it containing the point 10, 32 . 11. Resistance Due to a Conductor The resistance (in ohms) of a circular conductor varies directly with the length of the conductor and inversely with the square of the radius of the conductor. If 50 feet of wire with a radius of 6 * 10-3 inch has a resistance of 10 ohms, what would be the resistance of 100 feet of the same wire if the radius were increased to 7 * 10-3 inch? Cumulative Review In Problems 1–8, find the real solution(s), if any, of each equation. 1. 3x - 5 = 0 2. x - x - 12 = 0 3. 2x - 5x - 3 = 0 4. x2 - 2x - 2 = 0 5. x2 + 2x + 5 = 0 6. 22x + 1 = 3 2 7. 0 x - 2 0 = 1 2 8. 2x2 + 4x = 2 In Problems 9 and 10, solve each equation in the complex number system. 9. x2 = - 9 10. x2 - 2x + 5 = 0 In Problems 11–14, solve each inequality. Graph the solution set. 11. 2x - 3 … 7 13. 0 x - 2 0 … 1 12. - 1 6 x + 4 6 5 14. 0 2 + x 0 7 3 Chapter Project 15. Find the distance between the points P = 1 - 1, 32 and Q = 14, - 22 . Find the midpoint of the line segment from P to Q. 16. Which of the following points are on the graph of y = x3 - 3x + 1? (a) 1 - 2, - 12 (b) 12, 32 (c) 13, 12 17. Sketch the graph of y = x3. 197 18. Find the equation of the line containing the points 1 - 1, 42 and 12, - 22 . Express your answer in slope–intercept form. 19. Find the equation of the line perpendicular to the line y = 2x + 1 and containing the point 13, 52 . Express your answer in slope–intercept form and graph the line. 20. Graph the equation x2 + y2 - 4x + 8y - 5 = 0. Chapter Project The graph below, called a scatter diagram, shows the points (291.5, 268), (320, 305), c, (368, 385) in a Cartesian plane. From the graph, it appears that the data follow a linear relation. Sale Price ($ thousands) Zestimate vs. Sale Price in Oak Park, IL Internet-based Project Determining the Selling Price of a Home Determining how much to pay for a home is one of the more difficult decisions that must be made when purchasing a home. There are many factors that play a role in a home’s value. Location, size, number of bedrooms, number of bathrooms, lot size, and building materials are just a few. Fortunately, the website Zillow.com has developed its own formula for predicting the selling price of a home. This information is a great tool for predicting the actual sale price. For example, the data below show the “zestimate”—the selling price of a home as predicted by the folks at Zillow—and the actual selling price of the home, for homes in Oak Park, Illinois. 380 360 340 320 300 280 300 320 340 360 Zestimate ($ thousands) 1. Imagine drawing a line through the data that appears to fit the data well. Do you believe the slope of the line would be positive, negative, or close to zero? Why? 2. Pick two points from the scatter diagram. Treat the zestimate as the value of x, and treat the sale price as the corresponding value of y. Find the equation of the line through the two points you selected. 3. Interpret the slope of the line. 4. Use your equation to predict the selling price of a home whose zestimate is $335,000. Zestimate ($ thousands) Sale Price ($ thousands) 291.5 268 320 305 371.5 375 303.5 283 351.5 350 314 275 (b) Select two points from the scatter diagram and find the equation of the line through the points. 332.5 356 (c) Interpret the slope. 295 300 313 285 368 385 (d) Find a home from the Zillow website that interests you under the “Make Me Move” option for which a zestimate is available. Use your equation to predict the sale price based on the zestimate. 5. Do you believe it would be a good idea to use the equation you found in part 2 if the zestimate were $950,000? Why or why not? 6. Choose a location in which you would like to live. Go to www.zillow.com and randomly select at least ten homes that have recently sold. (a) Draw a scatter diagram of your data. 3 Functions and Their Graphs Choosing a Wireless Data Plan Most consumers choose a cellular provider first and then select an appropriate data plan from that provider. The choice as to the type of plan selected depends on your use of the device. For example, is online gaming important? Do you want to stream audio or video? The mathematics learned in this chapter can help you decide what plan is best suited to your particular needs. —See the Internet-based Chapter Project— Outline 3.1 3.2 3.3 3.4 3.5 3.6 198 Functions The Graph of a Function Properties of Functions Library of Functions; Piecewise-defined Functions Graphing Techniques: Transformations Mathematical Models: Building Functions Chapter Review Chapter Test Cumulative Review Chapter Projects A Look Back So far, our discussion has focused on techniques for graphing equations containing two variables. A Look Ahead In this chapter, we look at a special type of equation involving two variables called a function. This chapter deals with what a function is, how to graph functions, properties of functions, and how functions are used in applications. The word function apparently was introduced by René Descartes in 1637. For him, a function was simply any positive integral power of a variable x. Gottfried Wilhelm Leibniz (1646–1716), who always emphasized the geometric side of mathematics, used the word function to denote any quantity associated with a curve, such as the coordinates of a point on the curve. Leonhard Euler (1707–1783) employed the word to mean any equation or formula involving variables and constants. His idea of a function is similar to the one most often seen in courses that precede calculus. Later, the use of functions in investigating heat flow equations led to a very broad definition that originated with Lejeune Dirichlet (1805–1859), which describes a function as a correspondence between two sets. That is the definition used in this text. SECTION 3.1 Functions 199 3.1 Functions PREPARING FOR THIS SECTION Before getting started, review the following: r Intervals (Section 1.5, pp. 120–121) r Solving Inequalities (Section 1.5, pp. 123–126) r Evaluating Algebraic Expressions, Domain of a Variable (Chapter R, Section R.2, pp. 20–23) r Rationalizing Denominators (Chapter R, Section R.8, p. 75) Now Work the ‘Are You Prepared?’ problems on page 210. OBJECTIVES 1 2 3 4 5 1 Determine Whether a Relation Represents a Function y 5 (1, 2) 24 22 (0, 1) Determine Whether a Relation Represents a Function (p. 199) Find the Value of a Function (p. 202) Find the Difference Quotient of a Function (p. 205) Find the Domain of a Function Defined by an Equation (p. 206) Form the Sum, Difference, Product, and Quotient of Two Functions (p. 208) 2 4 x 25 Figure 1 y = 3x - 1 EXAMPL E 1 Often there are situations where the value of one variable is somehow linked to the value of another variable. For example, an individual’s level of education is linked to annual income. Engine size is linked to gas mileage. When the value of one variable is related to the value of a second variable, we have a relation. A relation is a correspondence between two sets. If x and y are two elements, one from each of these sets, and if a relation exists between x and y, then we say that x corresponds to y or that y depends on x, and we write x S y. There are a number of ways to express relations between two sets. For example, the equation y = 3x - 1 shows a relation between x and y. It says that if we take some number x, multiply it by 3, and then subtract 1, we obtain the corresponding value of y. In this sense, x serves as the input to the relation, and y is the output of the relation. This relation, expressed as a graph, is shown in Figure 1. The set of all inputs for a relation is called the domain of the relation, and the set of all outputs is called the range. In addition to being expressed as equations and graphs, relations can be expressed through a technique called mapping. A map illustrates a relation as a set of inputs with an arrow drawn from each element in the set of inputs to the corresponding element in the set of outputs. Ordered pairs can be used to represent x S y as 1x, y2. Maps and Ordered Pairs as Relations Figure 2 shows a relation between states and the number of representatives each state has in the House of Representatives. (Source: www.house.gov). The relation might be named “number of representatives.” State Number of Representatives Alaska Arizona California Colorado Florida North Dakota 1 7 9 27 53 Figure 2 Number of representatives In this relation, Alaska corresponds to 1, Arizona corresponds to 9, and so on. Using ordered pairs, this relation would be expressed as 5 1Alaska, 12, 1Arizona, 92, 1California, 532, 1Colorado, 72, 1Florida, 272, 1North Dakota, 12 6 200 CHAPTER 3 Functions and Their Graphs Person Phone Number Dan 555 – 2345 Gizmo 549 – 9402 930 – 3956 Colleen 555 – 8294 Phoebe 839 – 9013 Figure 3 Phone numbers Animal Life Expectancy Dog 11 Duck 10 Kangaroo Rabbit 7 Figure 4 Animal life expectancy DEFINITION y x X Domain Range Y The domain of the relation is {Alaska, Arizona, California, Colorado, Florida, North Dakota}, and the range is {1, 7, 9, 27, 53}. Note that the output “1” is listed only once in the range. r One of the most important concepts in algebra is the function. A function is a special type of relation. To understand the idea behind a function, let’s revisit the relation presented in Example 1. If we were to ask, “How many representatives does Alaska have?” you would respond “1.” In fact, each input state corresponds to a single output number of representatives. Let’s consider a second relation, one that involves a correspondence between four people and their phone numbers. See Figure 3. Notice that Colleen has two telephone numbers. There is no single answer to the question “What is Colleen’s phone number?” Let’s look at one more relation. Figure 4 is a relation that shows a correspondence between type of animal and life expectancy. If asked to determine the life expectancy of a dog, we would all respond, “11 years.” If asked to determine the life expectancy of a rabbit, we would all respond, “7 years.” Notice that the relations presented in Figures 2 and 4 have something in common. What is it? In both of these relations, each input corresponds to exactly one output. This leads to the definition of a function. Let X and Y be two nonempty sets.* A function from X into Y is a relation that associates with each element of X exactly one element of Y. The set X is called the domain of the function. For each element x in X, the corresponding element y in Y is called the value of the function at x, or the image of x. The set of all images of the elements in the domain is called the range of the function. See Figure 5. Since there may be some elements in Y that are not the image of some x in X, it follows that the range of a function may be a subset of Y, as shown in Figure 5. Not all relations between two sets are functions. The next example shows how to determine whether a relation is a function. Figure 5 EX AMPLE 2 Determining Whether a Relation Is a Function For each relation in Figures 6, 7, and 8, state the domain and range. Then determine whether the relation is a function. (a) See Figure 6. For this relation, the input is the number of calories in a fast-food sandwich, and the output is the fat content (in grams). Calories Fat (Wendy’s 1/4-lb Single) 580 31 (Burger King Whopper) 650 37 (Culver’s Deluxe Single) 541 33 (McDonald's Big Mac) 550 29 (Five Guys Hamburger) 700 43 Figure 6 Fat content Source: Each company’s Website *The sets X and Y will usually be sets of real numbers, in which case a (real) function results. The two sets can also be sets of complex numbers, and then we have defined a complex function. In the broad definition (proposed by Lejeune Dirichlet), X and Y can be any two sets. SECTION 3.1 Functions 201 (b) See Figure 7. For this relation, the inputs are gasoline stations in Harris County, Texas, and the outputs are the price per gallon of unleaded regular in March 2014. (c) See Figure 8. For this relation, the inputs are the weight (in carats) of pear-cut diamonds and the outputs are the price (in dollars). Gas Station Price per Gallon Valero $3.19 Shell $3.29 Texaco $3.35 Carats Price 0.70 $1529 0.71 $1575 0.75 $1765 0.78 $1798 Citgo $1952 Figure 7 Unleaded price per gallon Figure 8 Diamond price Source: Used with permission of Diamonds.com Solution (a) The domain of the relation is {541, 550, 580, 650, 700}, and the range of the relation is {29, 31, 33, 37, 43}. The relation in Figure 6 is a function because each element in the domain corresponds to exactly one element in the range. (b) The domain of the relation is {Citgo, Shell, Texaco, Valero}. The range of the relation is {$3.19, $3.29, $3.35}. The relation in Figure 7 is a function because each element in the domain corresponds to exactly one element in the range. Notice that it is okay for more than one element in the domain to correspond to the same element in the range (Shell and Citgo both sell gas for $3.29 a gallon). (c) The domain of the relation is {0.70, 0.71, 0.75, 0.78} and the range is {$1529, $1575, $1765, $1798, $1952}. The relation in Figure 8 is not a function because not every element in the domain corresponds to exactly one element in the range. If a 0.71-carat diamond is chosen from the domain, a single price cannot be assigned to it. r Now Work In Words For a function, no input has more than one output. The domain of a function is the set of all inputs; the range is the set of all outputs. EXAMPL E 3 PROBLEM 19 The idea behind a function is its predictability. If the input is known, we can use the function to determine the output. With “nonfunctions,” we don’t have this predictability. Look back at Figure 6. If asked, “How many grams of fat are in a 580-calorie sandwich?” we could use the correspondence to answer, “31.” Now consider Figure 8. If asked, “What is the price of a 0.71-carat diamond?” we could not give a single response because two outputs result from the single input “0.71.” For this reason, the relation in Figure 8 is not a function. We may also think of a function as a set of ordered pairs 1x, y2 in which no ordered pairs have the same first element and different second elements. The set of all first elements x is the domain of the function, and the set of all second elements y is its range. Each element x in the domain corresponds to exactly one element y in the range. Determining Whether a Relation Is a Function For each relation, state the domain and range. Then determine whether the relation is a function. (a) 5 11, 42, 12, 52, 13, 62, 14, 72 6 (b) 5 11, 42, 12, 42, 13, 52, 16, 102 6 (c) 5 1 - 3, 92, 1 - 2, 42, 10, 02, 11, 12 , 1 - 3, 82 6 202 CHAPTER 3 Functions and Their Graphs Solution (a) The domain of this relation is {1, 2, 3, 4}, and its range is {4, 5, 6, 7}. This relation is a function because there are no ordered pairs with the same first element and different second elements. (b) The domain of this relation is {1, 2, 3, 6}, and its range is {4, 5, 10}. This relation is a function because there are no ordered pairs with the same first element and different second elements. (c) The domain of this relation is { - 3, - 2, 0, 1}, and its range is {0, 1, 4, 8, 9}. This relation is not a function because there are two ordered pairs, 1 - 3, 92 and 1 - 3, 82, that have the same first element and different second elements. r In Example 3(b), notice that 1 and 2 in the domain both have the same image in the range. This does not violate the definition of a function; two different first elements can have the same second element. A violation of the definition occurs when two ordered pairs have the same first element and different second elements, as in Example 3(c). Now Work PROBLEM 23 Up to now we have shown how to identify when a relation is a function for relations defined by mappings (Example 2) and ordered pairs (Example 3). But relations can also be expressed as equations. The circumstances under which equations are functions are discussed next. To determine whether an equation, where y depends on x, is a function, it is often easiest to solve the equation for y. If any value of x in the domain corresponds to more than one y, the equation does not define a function; otherwise, it does define a function. EX AMPLE 4 Determining Whether an Equation Is a Function Determine whether the equation y = 2x - 5 defines y as a function of x. Solution The equation tells us to take an input x, multiply it by 2, and then subtract 5. For any input x, these operations yield only one output y, so the equation is a function. For example, if x = 1, then y = 2112 - 5 = - 3. If x = 3, then y = 2132 - 5 = 1. The graph of the equation y = 2x - 5 is a line with slope 2 and y-intercept - 5. The function is called a linear function. r EX AMPLE 5 Determining Whether an Equation Is a Function Determine whether the equation x2 + y2 = 1 defines y as a function of x. Solution To determine whether the equation x2 + y2 = 1, which defines the unit circle, is a function, solve the equation for y. x2 + y 2 = 1 y 2 = 1 - x2 y = { 21 - x2 For values of x for which - 1 6 x 6 1, two values of y result. For example, if x = 0, then y = {1, so two different outputs result from the same input. This means that the equation x2 + y2 = 1 does not define a function. r Now Work PROBLEM 37 2 Find the Value of a Function Functions are often denoted by letters such as f, F, g, G, and others. If f is a function, then for each number x in its domain, the corresponding image in the range is designated by the symbol f1x2, read as “ f of x” or as “ f at x.”We refer to f1x2 as the value of f at the number x ; f1x2 is the number that results when x is given and the function f is applied; f1x2 is the output corresponding to x or f1x2 is the image of x; f1x2 SECTION 3.1 Functions 203 does not mean “ f times x.” For example, the function given in Example 4 may be written as y = f1x2 = 2x - 5. Then f112 = - 3 and f132 = 1. Figure 9 illustrates some other functions. Notice that in every function, for each x in the domain, there is one value in the range. 1 f (1) f (1) 1 1 0 0 f (0) 2 2 f ( 2) x 2 1–2 F(2) 1 1 F(1) f(x ) x 2 Domain Range F(4) 1 F(x ) – x x Domain Range 1 (b) F (x ) – x (a) f (x ) x 2 0 0 g(0) 0 1 1 g(1) 2 2 g(2) 2 1– 4 4 3 G(0) G(2) G(3) 3 2 g(4) 4 x g(x) x Domain Range G(x) 3 x Domain Range (d) G (x) 3 (c) g(x) x Figure 9 Sometimes it is helpful to think of a function f as a machine that receives as input a number from the domain, manipulates it, and outputs a value. See Figure 10. The restrictions on this input/output machine are as follows: Input x x f 1. It accepts only numbers from the domain of the function. 2. For each input, there is exactly one output (which may be repeated for different inputs). Output y f(x) Figure 10 Input/output machine EXAMPL E 6 For a function y = f1x2, the variable x is called the independent variable, because it can be assigned any of the permissible numbers from the domain. The variable y is called the dependent variable, because its value depends on x. Any symbols can be used to represent the independent and dependent variables. For example, if f is the cube function, then f can be given by f1x2 = x3 or f1t2 = t 3 or f1z2 = z3. All three functions are the same. Each says to cube the independent variable to get the output. In practice, the symbols used for the independent and dependent variables are based on common usage, such as using C for cost in business. The independent variable is also called the argument of the function. Thinking of the independent variable as an argument can sometimes make it easier to find the value of a function. For example, if f is the function defined by f1x2 = x3, then f tells us to cube the argument. Thus f122 means to cube 2, f1a2 means to cube the number a, and f1x + h2 means to cube the quantity x + h. Finding Values of a Function For the function f defined by f1x2 = 2x2 - 3x, evaluate (a) f132 (e) - f1x2 (b) f1x2 + f 132 (f) f13x2 (c) 3f1x2 (g) f1x + 32 (d) f1 - x2 204 CHAPTER 3 Functions and Their Graphs Solution (a) Substitute 3 for x in the equation for f , f(x) = 2x2 - 3x, to get f 132 = 2132 2 - 3132 = 18 - 9 = 9 The image of 3 is 9. (b) f1x2 + f 132 = 12x2 - 3x2 + 192 = 2x2 - 3x + 9 (c) Multiply the equation for f by 3. 3f1x2 = 312x2 - 3x2 = 6x2 - 9x (d) Substitute - x for x in the equation for f and simplify. f1 - x2 = 21 - x2 2 - 31 - x2 = 2x2 + 3x Notice the use of parentheses here. (e) - f1x2 = - 12x2 - 3x2 = - 2x2 + 3x (f) Substitute 3x for x in the equation for f and simplify. f 13x2 = 213x2 2 - 313x2 = 219x2 2 - 9x = 18x2 - 9x (g) Substitute x + 3 for x in the equation for f and simplify. f1x + 32 = 21x + 32 2 - 31x + 32 = 21x2 + 6x + 92 - 3x - 9 = 2x2 + 12x + 18 - 3x - 9 = 2x2 + 9x + 9 r Notice in this example that f 1x + 32 ≠ f 1x2 + f132 , f 1 - x2 ≠ - f1x2 , and 3f1x2 ≠ f13x2. Now Work PROBLEM 43 Most calculators have special keys that allow you to find the value of certain commonly used functions. For example, you should be able to find the square function f1x2 = x2, the square root function f1x2 = 1x, the reciprocal function 1 f1x2 = = x -1, and many others that will be discussed later in this text (such as x ln x and log x). Verify the results of Example 7, which follows, on your calculator. EX AMPLE 7 Finding Values of a Function on a Calculator (a) f1x2 = x2 1 (b) F1x2 = x (c) g1x2 = 1x f11.2342 = 1.2342 = 1.522756 1 F11.2342 = ≈ 0.8103727715 1.234 g11.2342 = 11.234 ≈ 1.110855526 r COMMENT Graphing calculators can be used to evaluate any function. Figure 11 shows the result obtained in Example 6(a) on a TI-84 Plus C graphing calculator with the function to be evaluated, f (x) = 2x 2 - 3x, in Y1. Figure 11 Evaluating f (x) = 2x2 - 3x for x = 3 ■ SECTION 3.1 Functions 205 Implicit Form of a Function In general, when a function f is defined by an equation in x and y, we say that the function f is given implicitly. If it is possible to solve the equation for y in terms of x, then we write y = f 1x2 and say that the function is given explicitly. For example, COMMENT The explicit form of a function is the form required by a graphing calculator. ■ Implicit Form 3x + y = 5 Explicit Form y = f 1x2 = - 3x + 5 x2 - y = 6 y = f1x2 = x2 - 6 4 y = f1x2 = x xy = 4 SUMMARY Important Facts about Functions (a) For each x in the domain of a function f, there is exactly one image f 1x2 in the range; however, an element in the range can result from more than one x in the domain. (b) f is the symbol that we use to denote the function. It is symbolic of the equation (rule) that we use to get from an x in the domain to f1x2 in the range. (c) If y = f 1x2, then x is called the independent variable or argument of f, and y is called the dependent variable or the value of f at x. 3 Find the Difference Quotient of a Function An important concept in calculus involves looking at a certain quotient. For a given function y = f 1x2, the inputs x and x + h, h ≠ 0, result in the images f1x2 and f1x + h2. The quotient of their differences f1x + h2 - f1x2 f1x + h2 - f1x2 = 1x + h2 - x h with h ≠ 0, is called the difference quotient of f at x. DEFINITION The difference quotient of a function f at x is given by f1x + h2 - f1x2 h h ≠ 0 (1) The difference quotient is used in calculus to define the derivative, which leads to applications such as the velocity of an object and optimization of resources. When finding a difference quotient, it is necessary to simplify the expression in order to cancel the h in the denominator, as illustrated in the following example. EXAMPL E 8 Finding the Difference Quotient of a Function Find the difference quotient of each function. (a) f1x2 = 2x2 - 3x 4 (b) f1x2 = x (c) f1x2 = 1x 206 CHAPTER 3 Functions and Their Graphs Solution (a) f1x + h2 - f 1x2 3 21x + h2 2 - 31x + h2 4 - 3 2x2 - 3x4 = h h c f (x + h) = 2(x + h)2 - 3(x + h) = = = = = (b) f1x + h2 - f1x2 = h = = = = (c) 21x2 + 2xh + h2 2 - 3x - 3h - 2x2 + 3x Simplify. h 2x2 + 4xh + 2h2 - 3h - 2x2 Distribute and combine like terms. h 4xh + 2h2 - 3h Combine like terms. h h 14x + 2h - 32 Factor out h. h 4x + 2h - 3 Divide out the h’s. 4 4 x x + h 4 f (x + h) = x + h h 4x - 41x + h2 x1x + h2 Subtract. h 4x - 4x - 4h Divide and distribute. x1x + h2h - 4h Simplify. x1x + h2h 4 Divide out the factor h. x1x + h2 f1x + h2 - f1x2 2x + h - 2x = h h = = = = Now Work f (x + h) = 1x + h 2x + h - 2x # 2x + h + 2x h 2x + h + 2x 1 2x + h 2 2 - 1 2x 2 2 h 1 2x + h + 2x 2 h h 1 2x + h + 2x 2 1 2x + h + 2x PROBLEM 1 2x Rationalize the numerator. (A - B )(A + B ) = A2 - B 2 + h22 - 1 2x 2 2 = x + h - x = h Divide out the factor h. r 79 4 Find the Domain of a Function Defined by an Equation Often the domain of a function f is not specified; instead, only the equation defining the function is given. In such cases, we agree that the domain of f is the largest set of real numbers for which the value f 1x2 is a real number. The domain of a function f is the same as the domain of the variable x in the expression f1x2. EX AMPLE 9 Finding the Domain of a Function Find the domain of each of the following functions. (a) f1x2 = x2 + 5x (b) g1x2 = 3x x - 4 (c) h 1t2 = 24 - 3t (d) F(x) = 23x + 12 x - 5 2 SECTION 3.1 Functions Solution In Words The domain of g found in Example 9(b) is 5 x x ≠ - 2, x ≠ 2 6 . This notation is read, “The domain of the function g is the set of all real numbers x such that x does not equal - 2 and x does not equal 2.” 207 (a) The function says to square a number and then add five times the number. Since these operations can be performed on any real number, the domain of f is the set of all real numbers. (b) The function g says to divide 3x by x2 - 4. Since division by 0 is not defined, the denominator x2 - 4 can never be 0, so x can never equal - 2 or 2. The domain of the function g is 5 x x ≠ - 2, x ≠ 26 . (c) The function h says to take the square root of 4 - 3t. But only nonnegative numbers have real square roots, so the expression under the square root (the radicand) must be nonnegative (greater than or equal to zero). This requires that 4 - 3t Ú 0 - 3t Ú - 4 4 t … 3 4 4 f , or the interval a - q , d . 3 3 (d) The function F says to take the square root of 3x + 12 and divide this result by x - 5. This requires that 3x + 12 Ú 0, so x Ú - 4, and also that x - 5 ≠ 0, so x ≠ 5. Combining these two restrictions, the domain of F is The domain of h is e t ` t … 5 x x Ú - 4, x ≠ 56 . r The following steps may prove helpful for finding the domain of a function that is defined by an equation and whose domain is a subset of the real numbers. Finding the Domain of a Function Defined by an Equation 1. Start with the domain as the set of real numbers. 2. If the equation has a denominator, exclude any numbers that give a zero denominator. 3. If the equation has a radical of even index, exclude any numbers that cause the expression inside the radical (the radicand) to be negative. Now Work PROBLEM 55 If x is in the domain of a function f, we shall say that f is defined at x, or f(x) exists. If x is not in the domain of f, we say that f is not defined at x, or f(x) does not x exist. For example, if f 1x2 = 2 , then f102 exists, but f112 and f1 - 12 do not x - 1 exist. (Do you see why?) We have not said much about finding the range of a function. We will say more about finding the range when we look at the graph of a function in the next section. When a function is defined by an equation, it can be difficult to find the range. Therefore, we shall usually be content to find just the domain of a function when the function is defined by an equation. We shall express the domain of a function using inequalities, interval notation, set notation, or words, whichever is most convenient. When we use functions in applications, the domain may be restricted by physical or geometric considerations. For example, the domain of the function f defined by f 1x2 = x2 is the set of all real numbers. However, if f is used to obtain the area of a square when the length x of a side is known, then we must restrict the domain of f to the positive real numbers, since the length of a side can never be 0 or negative. 208 CHAPTER 3 Functions and Their Graphs EX A MPL E 1 0 Finding the Domain in an Application Express the area of a circle as a function of its radius. Find the domain. Solution A r Figure 12 Circle of radius r See Figure 12. The formula for the area A of a circle of radius r is A = pr 2. Using r to represent the independent variable and A to represent the dependent variable, the function expressing this relationship is A 1r2 = pr 2 In this setting, the domain is 5 r r 7 06 . (Do you see why?) r Observe, in the solution to Example 10, that the symbol A is used in two ways: It is used to name the function, and it is used to symbolize the dependent variable. This double use is common in applications and should not cause any difficulty. Now Work PROBLEM 97 5 Form the Sum, Difference, Product, and Quotient of Two Functions Next we introduce some operations on functions. Functions, like numbers, can be added, subtracted, multiplied, and divided. For example, if f1x2 = x2 + 9 and g1x2 = 3x + 5, then f1x2 + g1x2 = 1x2 + 92 + 13x + 52 = x2 + 3x + 14 The new function y = x2 + 3x + 14 is called the sum function f + g. Similarly, f1x2 # g1x2 = 1x2 + 92 13x + 52 = 3x3 + 5x2 + 27x + 45 The new function y = 3x3 + 5x2 + 27x + 45 is called the product function f # g. The general definitions are given next. DEFINITION If f and g are functions: The sum f + g is the function defined by 1f + g2 1x2 = f1x2 + g1x2 In Words Remember, the symbol x stands for intersection. It means you should find the elements that are common to two sets. DEFINITION The domain of f + g consists of the numbers x that are in the domains of both f and g. That is, domain of f + g = domain of f ∩ domain of g. The difference f − g is the function defined by 1f - g2 1x2 = f1x2 - g1x2 The domain of f - g consists of the numbers x that are in the domains of both f and g. That is, domain of f - g = domain of f ∩ domain of g. DEFINITION The product f ~ g is the function defined by 1f # g2 1x2 = f1x2 # g1x2 The domain of f # g consists of the numbers x that are in the domains of both f and g. That is, domain of f # g = domain of f ∩ domain of g. SECTION 3.1 Functions DEFINITION The quotient 209 f is the function defined by g f1x2 f a b 1x2 = g g1x2 g1x2 ≠ 0 f consists of the numbers x for which g1x2 ≠ 0 and that are in g the domains of both f and g. That is, The domain of domain of EX AM PL E 11 f = {x 0 g(x) ≠ 0} ∩ domain of f ∩ domain of g g Operations on Functions Let f and g be two functions defined as 1 x and g1x2 = x + 2 x - 1 Find the following functions, and determine the domain in each case. f (a) 1f + g2 1x2 (b) 1f - g2 1x2 (c) 1f # g2 1x2 (d) a b 1x2 g f1x2 = Solution The domain of f is 5 x x ≠ - 26 and the domain of g is 5 x x ≠ 16 . (a) 1f + g2 1x2 = f 1x2 + g1x2 = 1 x + x + 2 x - 1 x1x + 22 x - 1 x2 + 3x - 1 = + = 1x + 22 1x - 12 1x + 22 1x - 12 1x + 22 1x - 12 The domain of f + g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f + g is 5 x x ≠ - 2, x ≠ 16 . (b) 1f - g2 1x2 = f1x2 - g1x2 = 1 x x + 2 x - 1 x1x - 1x2 + x + 12 + 22 x - 1 = = 1x + 22 1x - 12 1x + 22 1x - 12 1x + 22 1x - 12 The domain of f - g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f - g is 5 x x ≠ - 2, x ≠ 16 . (c) 1f # g2 1x2 = f 1x2 # g1x2 = 1 # x x = x + 2 x - 1 1x + 22 1x - 12 The domain of f # g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f # g is 5 x x ≠ - 2, x ≠ 16 . 1 f 1x2 f x + 2 1 #x - 1 x - 1 (d) a b 1x2 = = = = g x x g1x2 x + 2 x1x + 22 x - 1 f The domain of consists of the numbers x for which g1x2 ≠ 0 and that are in g the domains of both f and g. Since g1x2 = 0 when x = 0, we exclude 0 as well f as - 2 and 1 from the domain. The domain of is 5 x x ≠ - 2, x ≠ 0, x ≠ 16 . g Now Work PROBLEM 67 r 210 CHAPTER 3 Functions and Their Graphs In calculus, it is sometimes helpful to view a complicated function as the sum, difference, product, or quotient of simpler functions. For example, F1x2 = x2 + 1x is the sum of f1x2 = x2 and g1x2 = 1x. H1x2 = x2 - 1 x2 + 1 is the quotient of f1x2 = x2 - 1 and g1x2 = x2 + 1. SUMMARY Function A relation between two sets of real numbers so that each number x in the first set, the domain, has corresponding to it exactly one number y in the second set. A set of ordered pairs 1x, y2 or 1x, f1x2 2 in which no first element is paired with two different second elements. The range is the set of y-values of the function that are the images of the x-values in the domain. A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f1x2 . Unspecified domain If a function f is defined by an equation and no domain is specified, then the domain will be taken to be the largest set of real numbers for which the equation defines a real number. Function notation y = f1x2 f is a symbol for the function. x is the independent variable, or argument. y is the dependent variable. f1x2 is the value of the function at x, or the image of x. 3.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. The inequality - 1 6 x 6 3 can be written in interval notation as . (pp. 120–121) 2. If x = - 2, the value of the 1 .(pp. 20–23) expression 3x2 - 5x + is x x - 3 is 3. The domain of the variable in the expression x + 4 . (pp. 20–23) 4. Solve the inequality: 3 - 2x 7 5. Graph the solution set. (pp. 123–126) 3 , multiply the 5. To rationalize the denominator of 25 - 2 numerator and denominator by (p. 75) 6. A quotient is considered rationalized if its denominator . (p. 75) contains no Concepts and Vocabulary 7. If f is a function defined by the equation y = f 1x2, then x is called the variable, and y is the variable. 8. If the domain of f is all real numbers in the interval 30, 74 , and the domain of g is all real numbers in the interval 3 - 2, 54, then the domain of f + g is all real numbers in the interval . f 9. The domain of consists of numbers x for which g1x2 0 g and . that are in the domains of both 10. If f 1x2 = x + 1 and g1x2 = x3, then = x3 - 1x + 12. 11. True or False Every relation is a function. 12. True or False The domain of 1f # g2 1x2 consists of the numbers x that are in the domains of both f and g. 13. True or False If no domain is specified for a function f, then the domain of f is taken to be the set of real numbers. 14. True or False The domain of the function f 1x2 = is 5x x ≠ {26 . x2 - 4 x 15. The set of all images of the elements in the domain of a function is called the . (a) range (b) domain (c) solution set (d) function 16. The independent variable is sometimes referred to as the of the function. (a) range (b) value (c) argument (d) definition f(x + h) - f(x) is called the of f. 17. The expression h (a) radicand (b) image (c) correspondence (d) difference quotient 18. When written as y = f(x), a function is said to be defined . (a) explicitly (b) consistently (c) implicitly (d) rationally SECTION 3.1 Functions 211 Skill Building In Problems 19–30, state the domain and range for each relation. Then determine whether each relation represents a function. 19. Person 20. Birthday Elvis Daughter Father Jan. 8 Bob Kaleigh Mar. 15 John Linda Marissa Sept. 17 Chuck Marcia Beth Colleen Diane 21. Hours Worked 22. Salary 20 Hours $200 $300 30 Hours $350 40 Hours $425 23. 5 12, 62, 1 - 3, 62, 14, 92, 12, 102 6 26. 5 10, - 22, 11, 32, 12, 32, 13, 72 6 29. 5 1 - 2, 42, 1 - 1, 12, 10, 02, 11, 12 6 Level of Education Average Income Less than 9th grade 9th-12th grade High School Graduate Some College College Graduate $18,120 $23,251 $36,055 $45,810 $67,165 24. 5 1 - 2, 52, 1 - 1, 32, 13, 72, 14, 122 6 25. 5 11, 32, 12, 32, 13, 32, 14, 32 6 27. 5 1 - 2, 42, 1 - 2, 62, 10, 32, 13, 72 6 28. 5 1 - 4, 42, 1 - 3, 32, 1 - 2, 22, 1 - 1, 12, 1 - 4, 02 6 30. 5 1 - 2, 162, 1 - 1, 42, 10, 32, 11, 42 6 In Problems 31–42, determine whether the equation defines y as a function of x. 34. y = 0 x 0 1 x 31. y = 2x2 - 3x + 4 32. y = x3 33. y = 35. y2 = 4 - x2 36. y = { 21 - 2x 37. x = y2 38. x + y2 = 1 3 39. y = 2 x 40. y = 41. 2x2 + 3y2 = 1 42. x2 - 4y2 = 1 3x - 1 x + 2 In Problems 43–50, find the following for each function: (a) f 102 (b) f 112 (c) f 1 - 12 (d) f 1 - x2 (e) - f 1x2 (f) f 1x + 12 (g) f 12x2 (h) f 1x + h2 x2 - 1 x + 4 43. f 1x2 = 3x2 + 2x - 4 44. f 1x2 = - 2x2 + x - 1 x 45. f 1x2 = 2 x + 1 47. f 1x2 = 0 x 0 + 4 48. f 1x2 = 2x2 + x 49. f 1x2 = 2x + 1 3x - 5 52. f 1x2 = x2 + 2 53. f 1x2 = x x + 1 54. f 1x2 = 2x x2 - 4 57. F 1x2 = x - 2 x3 + x 58. G1x2 = 2 Ax - 1 62. f 1x2 = 46. f 1x2 = 50. f 1x2 = 1 - 1 1x + 22 2 In Problems 51–66, find the domain of each function. 51. f 1x2 = - 5x + 4 55. g1x2 = x x2 - 16 56. h1x2 = 2 59. h1x2 = 23x - 12 60. G1x2 = 21 - x 61. p1x2 = 63. f 1x2 = 64. f 1x2 = 65. P(t) = x 2x - 4 -x 2- x - 2 2t - 4 3t - 21 66. h(z) = In Problems 67–76, for the given functions f and g, find the following. For parts (a)–(d), also find the domain. f (b) 1f - g2 1x2 (c) 1f # g2 1x2 (d) a b 1x2 (a) 1f + g2 1x2 g f (e) 1f + g2 132 (f) 1f - g2 142 (g) 1f # g2 122 (h) a b 112 g 67. f 1x2 = 3x + 4; g1x2 = 2x - 3 69. f 1x2 = x - 1; g1x2 = 2x 2 68. f 1x2 = 2x + 1; g1x2 = 3x - 2 70. f 1x2 = 2x2 + 3; g1x2 = 4x3 + 1 x2 x + 1 2 x + 4 x3 - 4x 4 2x - 9 2z + 3 z - 2 212 CHAPTER 3 Functions and Their Graphs 71. f 1x2 = 1x; g1x2 = 3x - 5 72. f 1x2 = 0 x 0 ; g1x2 = x 73. f 1x2 = 1 + 74. f 1x2 = 2x - 1; g1x2 = 24 - x 1 1 ; g1x2 = x x 2x + 3 4x 75. f 1x2 = ; g1x2 = 3x - 2 3x - 2 76. f 1x2 = 2x + 1; g1x2 = 77. Given f 1x2 = 3x + 1 and 1f + g2 1x2 = 6 - 1 x, find the 2 78. Given f 1x2 = 2 x f x + 1 1 , find the function g. and a b 1x2 = 2 x g x - x function g. In Problems 79–90, find the difference quotient of f; that is, find f 1x + h2 - f 1x2 h , h ≠ 0, for each function. Be sure to simplify. 79. f 1x2 = 4x + 3 80. f 1x2 = - 3x + 1 81. f(x) = x2 - 4 83. f 1x2 = x2 - x + 4 84. f 1x2 = 3x2 - 2x + 6 85. f(x) = 87. f(x) = 2x x + 3 88. f(x) = 5x x - 4 82. f(x) = 3x2 + 2 86. f 1x2 = 1 x2 89. f(x) = 2x - 2 1 x + 3 90. f(x) = 2x + 1 [Hint: Rationalize the numerator.] Applications and Extensions 91. Given f(x) = x2 - 2x + 3, find the value(s) for x such that f(x) = 11. 3 5 92. Given f(x) = x - , find the value(s) for x such that 6 4 7 f(x) = - . 16 93. If f 1x2 = 2x + Ax + 4x - 5 and f 122 = 5, what is the value of A ? 3 2 94. If f 1x2 = 3x2 - Bx + 4 and f 1 - 12 = 12, what is the value of B ? 3x + 8 95. If f 1x2 = and f 102 = 2, what is the value of A ? 2x - A 96. If f 1x2 = 2x - B 1 and f 122 = , what is the value of B ? 3x + 4 2 represents the number N of housing units (in millions) in 2012 that had r rooms, where r is an integer and 2 … r … 9. (a) Identify the dependent and independent variables. (b) Evaluate N(3). Provide a verbal explanation of the meaning of N(3). 103. Effect of Gravity on Earth If a rock falls from a height of 20 meters on Earth, the height H (in meters) after x seconds is approximately H1x2 = 20 - 4.9x2 (a) What is the height of the rock when x = 1 second? x = 1.1 seconds? x = 1.2 seconds? x = 1.3 seconds? (b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters? (c) When does the rock strike the ground? 97. Geometry Express the area A of a rectangle as a function of the length x if the length of the rectangle is twice its width. 104. Effect of Gravity on Jupiter If a rock falls from a height of 20 meters on the planet Jupiter, its height H (in meters) after x seconds is approximately 98. Geometry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. H1x2 = 20 - 13x2 99. Constructing Functions Express the gross salary G of a person who earns $14 per hour as a function of the number x of hours worked. 100. Constructing Functions Tiffany, a commissioned salesperson, earns $100 base pay plus $10 per item sold. Express her gross salary G as a function of the number x of items sold. (a) What is the height of the rock when x = 1 second? x = 1.1 seconds? x = 1.2 seconds? (b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters? (c) When does the rock strike the ground? 101. Population as a Function of Age The function P 1a2 = 0.014a2 - 5.073a + 327.287 represents the population P (in millions) of Americans that are a years of age or older in 2012. (a) Identify the dependent and independent variables. (b) Evaluate P 1202 . Provide a verbal explanation of the meaning of P 1202 . (c) Evaluate P 102 . Provide a verbal explanation of the meaning of P 102 . Source: U.S. Census Bureau 102. Number of Rooms The function N1r2 = - 1.35r 2 + 15.45r - 20.71 105. Cost of Trans-Atlantic Travel A Boeing 747 crosses the Atlantic Ocean (3000 miles) with an airspeed of 500 miles per hour. The cost C (in dollars) per passenger is given by C 1x2 = 100 + 36,000 x + 10 x SECTION 3.1 Functions where x is the ground speed 1airspeed { wind2. (a) What is the cost per passenger for quiescent (no wind) conditions? (b) What is the cost per passenger with a head wind of 50 miles per hour? (c) What is the cost per passenger with a tail wind of 100 miles per hour? (d) What is the cost per passenger with a head wind of 100 miles per hour? 106. Cross-sectional Area The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A1x2 = 4x11 - x2 , where x represents the length, in feet, of half the base of the beam. See the figure. Determine the cross-sectional area of the beam if the length of half the base of the beam is as follows: (a) One-third of a foot A(x ) 4x 1 x 2 (b) One-half of a foot (c) Two-thirds of a foot 1 x 107. Economics The participation rate is the number of people in the labor force divided by the civilian population (excludes military). Let L 1x2 represent the size of the labor force in year x, and P 1x2 represent the civilian population in year x. Determine a function that represents the participation rate R as a function of x. 108. Crimes Suppose that V 1x2 represents the number of violent crimes committed in year x and P 1x2 represents the number of property crimes committed in year x. Determine a function T that represents the combined total of violent crimes and property crimes in year x. 109. Health Care Suppose that P 1x2 represents the percentage of income spent on health care in year x and I 1x2 represents income in year x. Determine a function H that represents total health care expenditures in year x. 213 110. Income Tax Suppose that I 1x2 represents the income of an individual in year x before taxes and T 1x2 represents the individual’s tax bill in year x. Determine a function N that represents the individual’s net income (income after taxes) in year x. 111. Profit Function Suppose that the revenue R, in dollars, from selling x cell phones, in hundreds, is R 1x2 = - 1.2x2 + 220x. The cost C, in dollars, of selling x cell phones, in hundreds, is C 1x2 = 0.05x3 - 2x2 + 65x + 500. (a) Find the profit function, P 1x2 = R 1x2 - C 1x2. (b) Find the profit if x = 15 hundred cell phones are sold. (c) Interpret P(15). 112. Profit Function Suppose that the revenue R, in dollars, from selling x clocks is R1x2 = 30x. The cost C, in dollars, of selling x clocks is C 1x2 = 0.1x2 + 7x + 400. (a) Find the profit function, P 1x2 = R 1x2 - C 1x2. (b) Find the profit if x = 30 clocks are sold. (c) Interpret P(30). 113. Stopping Distance When the driver of a vehicle observes an impediment, the total stopping distance involves both the reaction distance (the distance the vehicle travels while the driver moves his or her foot to the brake pedal) and the braking distance (the distance the vehicle travels once the brakes are applied). For a car traveling at a speed of v miles per hour, the reaction distance R, in feet, can be estimated by R(v) = 2.2v. Suppose that the braking distance B, in feet, for a car is given by B(v) = 0.05v2 + 0.4v - 15. (a) Find the stopping distance function D(v) = R(v) + B(v). (b) Find the stopping distance if the car is traveling at a speed of 60 mph. (c) Interpret D(60). 114. Some functions f have the property that f 1a + b2 = f 1a2 + f 1b2 for all real numbers a and b. Which of the following functions have this property? (a) h1x2 = 2x (b) g1x2 = x2 1 (c) F 1x2 = 5x - 2 (d) G1x2 = x Explaining Concepts: Discussion and Writing x2 - 1 the 115. Are the functions f 1x2 = x - 1 and g1x2 = x + 1 same? Explain. 117. Find a function H that multiplies a number x by 3 and then subtracts the cube of x and divides the result by your age. 116. Investigate when, historically, the use of the function notation y = f 1x2 first appeared. Retain Your Knowledge Problems 118–121 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 118. List the intercepts and test for symmetry: (x + 12)2 + y2 = 16 119. Determine which of the given points are on the graph of the equation. y = 3x2 - 82x 120. How many pounds of lean hamburger that is 7% fat must be mixed with 12 pounds of ground chuck that is 20% fat to have a hamburger mixture that is 15% fat? 121. Solve x3 - 9x = 2x2 - 18. Points: ( - 1, - 5), (4, 32), (9, 171) ‘Are You Prepared?’ Answers 1. 1 - 1, 32 2. 21.5 3. 5 x x ≠ - 46 4. 5x x 6 - 16 1 0 1 5. 25 + 2 6. radicals 214 CHAPTER 3 Functions and Their Graphs 3.2 The Graph of a Function PREPARING FOR THIS SECTION Before getting started, review the following: r Graphs of Equations (Section 2.2, pp. 157–159) r Intercepts (Section 2.2, pp. 159–160) Now Work the ‘Are You Prepared?’ problems on page 218. OBJECTIVES 1 Identify the Graph of a Function (p. 214) 2 Obtain Information from or about the Graph of a Function (p. 215) In applications, a graph often demonstrates more clearly the relationship between two variables than, say, an equation or table. For example, Table 1 shows the average price of gasoline in the United States for the years 1985–2014 (adjusted for inflation, based on 2014 dollars). If we plot these data and then connect the points, we obtain Figure 13. 1.80 2006 3.01 1987 1.89 1997 1.76 2007 3.19 1988 1.81 1998 1.49 2008 3.56 1989 1.87 1999 1.61 2009 2.58 1990 2.03 2000 2.03 2010 3.00 1991 1.91 2001 1.90 2011 3.69 1992 1.82 2002 1.76 2012 3.72 1993 1.74 2003 1.99 2013 3.54 1994 1.71 2004 2.31 2014 3.43 2013 2014 1996 2011 1.90 2009 1986 4.00 3.50 3.00 2.50 2.00 1.50 1.00 0.50 0.00 2007 2.74 2005 2005 2003 1.72 2001 1995 1999 2.55 1997 1985 1995 Price 1993 Year 1991 Price 1989 Year 1987 Price 1985 Year Price (dollars per gallon) Table 1 Figure 13 Average retail price of gasoline (2014 dollars) Source: U.S. Energy Information Administration Source: U.S. Energy Information Administration We can see from the graph that the price of gasoline (adjusted for inflation) stayed roughly the same from 1986 to 1991 and rose rapidly from 2002 to 2008. The graph also shows that the lowest price occurred in 1998. To learn information such as this from an equation requires that some calculations be made. Look again at Figure 13. The graph shows that for each date on the horizontal axis, there is only one price on the vertical axis. The graph represents a function, although the exact rule for getting from date to price is not given. When a function is defined by an equation in x and y, the graph of the function is the graph of the equation; that is, it is the set of points 1x, y2 in the xy-plane that satisfy the equation. 1 Identify the Graph of a Function In Words If any vertical line intersects a graph at more than one point, the graph is not the graph of a function. THEOREM Not every collection of points in the xy-plane represents the graph of a function. Remember, for a function, each number x in the domain has exactly one image y in the range. This means that the graph of a function cannot contain two points with the same x-coordinate and different y-coordinates. Therefore, the graph of a function must satisfy the following vertical-line test. Vertical-Line Test A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. SECTION 3.2 The Graph of a Function 215 Identifying the Graph of a Function EXAMPL E 1 Which of the graphs in Figure 14 are graphs of functions? y 6 y 4 y y 3 1 (1, 1) 4 3 (a) y x 2 Figure 14 Solution 3x 4x 4 (1, 1) 6 x 1 1 3 (b) y x 3 1 x (c) x y 2 (d) x 2 y 2 1 The graphs in Figures 14(a) and 14(b) are graphs of functions, because every vertical line intersects each graph in at most one point. The graphs in Figures 14(c) and 14(d) are not graphs of functions, because there is a vertical line that intersects each graph in more than one point. Notice in Figure 14(c) that the input 1 corresponds to two outputs, - 1 and 1. This is why the graph does not represent a function. r Now Work PROBLEM 17 2 Obtain Information from or about the Graph of a Function If 1x, y2 is a point on the graph of a function f, then y is the value of f at x; that is, y = f1x2 . Also if y = f1x2, then 1x, y2 is a point on the graph of f. For example, if 1 - 2, 72 is on the graph of f, then f 1 - 22 = 7, and if f 152 = 8, then the point 15, 82 is on the graph of y = f1x2 . The next example illustrates how to obtain information about a function if its graph is given. Obtaining Information from the Graph of a Function EXAMPL E 2 Let f be the function whose graph is given in Figure 15. (The graph of f might represent the distance y that the bob of a pendulum is from its at-rest position at time x. Negative values of y mean that the pendulum is to the left of the at-rest position, and positive values of y mean that the pendulum is to the right of the at-rest position.) y 4 2 (5––2, 0) (––2 , 0) (7––2, 0) (3––2, 0) 2 4 (4, 4) (2, 4) (0, 4) (, 4) x (3, 4) Figure 15 (a) What are f102, f a (b) (c) (d) (e) (f) (g) Solution 3p b , and f13p2? 2 What is the domain of f ? What is the range of f ? List the intercepts. (Recall that these are the points, if any, where the graph crosses or touches the coordinate axes.) How many times does the line y = 2 intersect the graph? For what values of x does f1x2 = - 4? For what values of x is f1x2 7 0? (a) Since 10, 42 is on the graph of f, the y-coordinate 4 is the value of f at the 3p x-coordinate 0; that is, f102 = 4. In a similar way, when x = , then y = 0, so 2 3p f a b = 0. When x = 3p, then y = - 4, so f 13p2 = - 4. 2 (b) To determine the domain of f, notice that the points on the graph of f have x-coordinates between 0 and 4p, inclusive; and for each number x between 0 and 4p, there is a point 1x, f1x2 2 on the graph. The domain of f is 5 x 0 … x … 4p6 or the interval 3 0, 4p4 . (c) The points on the graph all have y-coordinates between - 4 and 4, inclusive; and for each such number y, there is at least one number x in the domain. The range of f is 5 y - 4 … y … 46 or the interval 3 - 4, 44 . 216 CHAPTER 3 Functions and Their Graphs (d) The intercepts are the points p 3p 5p 10, 42, a , 0b , a , 0b , a , 0b , and 2 2 2 a 7p , 0b 2 (e) Draw the horizontal line y = 2 on the graph in Figure 15. Notice that the line intersects the graph four times. (f) Since 1p, - 42 and 13p, - 42 are the only points on the graph for which y = f1x2 = - 4, we have f1x2 = - 4 when x = p and x = 3p. (g) To determine where f1x2 7 0, look at Figure 15 and determine the x-values from 0 to 4p for which the y-coordinate is positive. This occurs p 3p 5p 7p b h a , b h a , 4p d . Using inequality notation, f1x2 7 0 2 2 2 2 p 3p 5p 7p for 0 … x 6 or 6 x 6 or 6 x … 4p. 2 2 2 2 on c 0, r When the graph of a function is given, its domain may be viewed as the shadow created by the graph on the x-axis by vertical beams of light. Its range can be viewed as the shadow created by the graph on the y-axis by horizontal beams of light. Try this technique with the graph given in Figure 15. Now Work EX AMPLE 3 PROBLEMS 11 AND 15 Obtaining Information about the Graph of a Function Consider the function: f1x2 = x + 1 x + 2 (a) Find the domain of f. 1 (b) Is the point a1, b on the graph of f ? 2 (c) If x = 2, what is f1x2? What point is on the graph of f ? (d) If f1x2 = 2, what is x? What point is on the graph of f ? (e) What are the x-intercepts of the graph of f (if any)? What point(s) are on the graph of f ? Solution (a) The domain of f is 5 x x ≠ - 26 . (b) When x = 1, then f112 = 2 1 + 1 = 1 + 2 3 f1x2 = x + 1 x + 2 1 2 The point a1, b is on the graph of f; the point a1, b is not. 3 2 (c) If x = 2, then f122 = 2 + 1 3 = 2 + 2 4 3 The point ¢ 2, ≤ is on the graph of f. 4 (d) If f1x2 = 2, then x x x x + + + + 1 2 1 1 x = 2 f1x2 = 2 = 21x + 22 = 2x + 4 = -3 Multiply both sides by x + 2. Distribute. Solve for x. If f1x2 = 2, then x = - 3. The point 1 - 3, 22 is on the graph of f. SECTION 3.2 The Graph of a Function 217 (e) The x-intercepts of the graph of f are the real solutions of the equation f1x2 = 0 that are in the domain of f. x + 1 = 0 x + 2 x + 1 = 0 x = -1 Multiply both sides by x + 2. Subtract 1 from both sides. x + 1 = 0 is x = - 1, so - 1 is x + 2 the only x-intercept. Since f1 - 12 = 0, the point ( - 1, 0) is on the graph of f. The only real solution of the equation f1x2 = r Now Work EXAMPL E 4 PROBLEM 27 Average Cost Function The average cost C per computer of manufacturing x computers per day is given by the function C 1x2 = 0.56x2 - 34.39x + 1212.57 + 20,000 x Determine the average cost of manufacturing: (a) 30 computers in a day (b) 40 computers in a day (c) 50 computers in a day (d) Graph the function C = C 1x2, 0 6 x … 80. (e) Create a TABLE with TblStart = 1 and ∆Tbl = 1. Which value of x minimizes the average cost? Solution (a) The average cost per computer of manufacturing x = 30 computers is C 1302 = 0.561302 2 - 34.391302 + 1212.57 + 20,000 = $1351.54 30 (b) The average cost per computer of manufacturing x = 40 computers is C 1402 = 0.561402 2 - 34.391402 + 1212.57 + 20,000 = $1232.97 40 (c) The average cost per computer of manufacturing x = 50 computers is C 1502 = 0.561502 2 - 34.391502 + 1212.57 + (d) See Figure 16 for the graph of C = C 1x2. 20,000 = $1293.07 50 (e) With the function C = C 1x2 in Y1 , we create Table 2. We scroll down until we find a value of x for which Y1 is smallest. Table 3 shows that manufacturing x = 41 computers minimizes the average cost at $1231.74 per computer. 4000 0 0 80 Figure 16 C1x2 = 0.56x2 - 34.39x + 1212.57 + 20,000 x Now Work Table 2 PROBLEM Table 3 35 r 218 CHAPTER 3 Functions and Their Graphs SUMMARY Graph of a Function The collection of points 1x, y2 that satisfies the equation y = f 1x2. Vertical-Line Test A collection of points is the graph of a function if and only if every vertical line intersects the graph in at most one point. 3.2 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. The intercepts of the equation x2 + 4y2 = 16 are (pp. 159–160) . 2. True or False The point 1 - 2, - 62 is on the graph of the equation x = 2y - 2. (pp. 157–159) Concepts and Vocabulary 3. A set of points in the xy-plane is the graph of a function if and only if every line intersects the graph in at most one point. 4. If the point 15, - 32 is a point on the graph of f, then f1 2 = . 5. Find a so that the point 1 - 1, 22 is on the graph of f 1x2 = ax2 + 4. 6. True or False Every graph represents a function. 7. True or False The graph of a function y = f 1x2 always crosses the y-axis. 8. True or False The y-intercept of the graph of the function y = f 1x2, whose domain is all real numbers, is f 102. 9. If a function is defined by an equation in x and y, then the set of points (x, y) in the xy-plane that satisfy the equation is called . (a) the domain of the function (b) the range of the function (c) the graph of the function (d) the relation of the function 10. The graph of a function y = f(x) can have more than one of which type of intercept? (a) x-intercept (b) y-intercept (c) both (d) neither Skill Building 11. Use the given graph of the function f to answer parts (a)–(n). y (0, 3) 4 12. Use the given graph of the function f to answer parts (a)–(n). y (2, 4) (4, 3) 4 (10, 0) (11, 1) (5, 3) (–4, 2) (–2, 1) 2 (–3, 0) (4, 0) 5 –5 (6, 0) (–5, –2) (–6, –3) –3 11 x (8, – 2) Find f 102 and f 1 - 62. Find f 162 and f 1112. Is f 132 positive or negative? Is f 1 - 42 positive or negative? For what values of x is f 1x2 = 0? For what values of x is f 1x2 7 0? What is the domain of f ? What is the range of f ? What are the x-intercepts? What is the y-intercept? 1 (k) How often does the line y = intersect the graph? 2 (l) How often does the line x = 5 intersect the graph? (m) For what values of x does f 1x2 = 3? (n) For what values of x does f 1x2 = - 2? (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) –4 –2 (0, 0) –2 2 4 (6, 0) 6 x (2, –2) (a) Find f 102 and f 162. (b) Find f 122 and f 1 - 22. (c) Is f 132 positive or negative? (d) Is f 1 - 12 positive or negative? (e) For what values of x is f 1x2 = 0? (f) For what values of x is f 1x2 6 0? (g) What is the domain of f ? (h) What is the range of f ? (i) What are the x-intercepts? (j) What is the y-intercept? (k) How often does the line y = - 1 intersect the graph? (l) How often does the line x = 1 intersect the graph? (m) For what value of x does f 1x2 = 3? (n) For what value of x does f 1x2 = - 2? SECTION 3.2 The Graph of a Function 219 In Problems 13–24, determine whether the graph is that of a function by using the vertical-line test. If it is, use the graph to find: (a) The domain and range (b) The intercepts, if any (c) Any symmetry with respect to the x-axis, the y-axis, or the origin 13. 14. y 3 3 3 3x 3 y (1, 2) 3 22. 3 3 23. y 3 3 26. f 1x2 = - 3x2 + 5x (a) Is the point 1 - 1, 22 on the graph of f ? (b) If x = - 2, what is f 1x2? What point is on the graph of f ? (c) If f 1x2 = - 2, what is x? What point(s) are on the graph of f ? (d) What is the domain of f ? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. x + 2 x - 6 Is the point 13, 142 on the graph of f ? If x = 4, what is f 1x2? What point is on the graph of f ? If f 1x2 = 2, what is x? What point(s) are on the graph of f ? What is the domain of f ? List the x-intercepts, if any, of the graph of f. List the y-intercept, if there is one, of the graph of f. 28. f 1x2 = x2 + 2 x + 4 20. y 4 (3, 2) 3 (a) Is the point a1, b on the graph of f ? 5 x (4, 3) 4 4x 4 24. y 3x 27. f 1x2 = –– 2 1 y (1–2 , 5) 4 6 3 –– 2 9 25. f 1x2 = 2x2 - x - 1 (a) Is the point 1 - 1, 22 on the graph of f ? (b) If x = - 2, what is f 1x2? What point is on the graph of f ? (c) If f 1x2 = - 1, what is x? What point(s) are on the graph of f ? (d) What is the domain of f ? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. (d) (e) (f) 3 In Problems 25–30, answer the questions about the given function. (a) (b) (c) x 3x x 3 –– 2 3 3x (1, 2) –– 2 1 1 y 3 3 3 21. 19. y 3 3 3x y 1 3x 18. y 3 16. y 3 3 17. 15. y 3 3 1 3 3 x (2, 3) 3 3 x (b) If x = 0, what is f 1x2? What point is on the graph of f ? 1 (c) If f 1x2 = , what is x? What point(s) are on the graph 2 of f ? (d) What is the domain of f ? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. 2x2 x4 + 1 Is the point 1 - 1, 12 on the graph of f ? If x = 2, what is f 1x2? What point is on the graph of f ? If f 1x2 = 1, what is x? What point(s) are on the graph of f ? What is the domain of f ? List the x-intercepts, if any, of the graph of f. List the y-intercept, if there is one, of the graph of f. 29. f 1x2 = (a) (b) (c) (d) (e) (f) 30. f 1x2 = 2x x - 2 2 1 (a) Is the point a , - b on the graph of f ? 2 3 (b) If x = 4, what is f 1x2? What point is on the graph of f ? (c) If f 1x2 = 1, what is x? What point(s) are on the graph of f ? (d) What is the domain of f ? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. 220 CHAPTER 3 Functions and Their Graphs Applications and Extensions 31. The graphs of two functions, f and g, are illustrated. Use the graphs to answer parts (a)–(f). y yg(x) (2, 2) 2 (2, 1) (4, 1) 2 2 (6, 1) (6, 0) x yf(x) 4 (3, 2) (5, 2) (4, 3) 4 (a) 1f + g2 122 (c) 1f - g2 162 (e) 1f # g2 122 (b) 1f + g2 142 (d) 1g - f2 162 f (f) a b 142 g 32. Granny Shots The last player in the NBA to use an underhand foul shot (a “granny” shot) was Hall of Fame forward Rick Barry, who retired in 1980. Barry believes that current NBA players could increase their free-throw percentage if they were to use an underhand shot. Since underhand shots are released from a lower position, the angle of the shot must be increased. If a player shoots an underhand foul shot, releasing the ball at a 70-degree angle from a position 3.5 feet above the floor, then the path of the ball can be modeled 136x2 by the function h1x2 = + 2.7x + 3.5, where h is v2 the height of the ball above the floor, x is the forward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second. (a) The center of the hoop is 10 feet above the floor and 15 feet in front of the foul line. Determine the initial velocity with which the ball must be shot in order for the ball to go through the hoop. (b) Write the function for the path of the ball using the velocity found in part (a). (c) Determine the height of the ball after it has traveled 9 feet in front of the foul line. (d) Find additional points and graph the path of the basketball. Source: The Physics of Foul Shots, Discover, Vol. 21, No. 10, October 2000 33. Free-throw Shots According to physicist Peter Brancazio, the key to a successful foul shot in basketball lies in the arc of the shot. Brancazio determined the optimal angle of the arc from the free-throw line to be 45 degrees. The arc also depends on the velocity with which the ball is shot. If a player shoots a foul shot, releasing the ball at a 45-degree angle from a position 6 feet above the floor, then the path of the ball can be modeled by the function h1x2 = - 44x2 + x + 6 v2 where h is the height of the ball above the floor, x is the forward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second. Suppose a player shoots a ball with an initial velocity of 28 feet per second. (a) Determine the height of the ball after it has traveled 8 feet in front of the foul line. (b) Determine the height of the ball after it has traveled 12 feet in front of the foul line. (c) Find additional points and graph the path of the basketball. (d) The center of the hoop is 10 feet above the floor and 15 feet in front of the foul line. Will the ball go through the hoop? Why or why not? If not, with what initial velocity must the ball be shot in order for the ball to go through the hoop? Source: The Physics of Foul Shots, Discover, Vol. 21, No. 10, October 2000 34. Cross-sectional Area The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A1x2 = 4x21 - x2 , where x represents the length, in feet, of half the base of the beam. See the figure. (a) Find the domain of A. (b) Use a graphing utility to graph the function A = A1x2. (c) Create a TABLE with TblStart = 0 and ∆Tbl = 0.1 for 0 … x … 1. Which value of x maximizes the crosssectional area? What should be the length of the base of the beam to maximize the cross-sectional area? A(x ) 4x 1 x 2 1 x 35. Motion of a Golf Ball A golf ball is hit with an initial velocity of 130 feet per second at an inclination of 45° to the horizontal. In physics, it is established that the height h of the golf ball is given by the function - 32x2 + x 1302 where x is the horizontal distance that the golf ball has traveled. h1x2 = SECTION 3.2 The Graph of a Function 2 4000 W 1h2 = ma b 4000 + h (a) If Amy weighs 120 pounds at sea level, how much will she weigh on Pikes Peak, which is 14,110 feet above sea level? (b) Use a graphing utility to graph the function W = W 1h2. Use m = 120 pounds. (c) Create a TABLE with TblStart = 0 and ∆Tbl = 0.5 to see how the weight W varies as h changes from 0 to 5 miles. (d) At what height will Amy weigh 119.95 pounds? (e) Does your answer to part (d) seem reasonable? Explain. 37. Cost of Trans-Atlantic Travel A Boeing 747 crosses the Atlantic Ocean (3000 miles) with an airspeed of 500 miles per hour. The cost C (in dollars) per passenger is given by C 1x2 = 100 + 36,000 x + 10 x where x is the groundspeed 1airspeed { wind2. (a) What is the cost when the groundspeed is 480 miles per hour? 600 miles per hour? (b) Find the domain of C. (c) Use a graphing utility to graph the function C = C 1x2. (d) Create a TABLE with TblStart = 0 and ∆Tbl = 50. (e) To the nearest 50 miles per hour, what groundspeed minimizes the cost per passenger? 38. Reading and Interpreting Graphs Let C be the function whose graph is given in the next column. This graph represents the cost C of manufacturing q computers in a day. (a) Determine C(0). Interpret this value. (b) Determine C(10). Interpret this value. (c) Determine C(50). Interpret this value. C (100, 280 000) 250,000 200,000 150,000 100,000 50,000 (50, 51 000) (30, 32 000) (10, 19 000) (0, 5000) 10 20 30 40 50 60 70 80 Number of computers 90 100 q 39. Reading and Interpreting Graphs Let C be the function whose graph is given below. This graph represents the cost C of using g gigabytes of data in a month for a shared-data family plan. (a) Determine C (0). Interpret this value. (b) Determine C (5). Interpret this value. (c) Determine C (15). Interpret this value. (d) What is the domain of C? What does this domain imply in terms of the number of gigabytes? (e) Describe the shape of the graph. C 480 (60, 405) Cost (dollars) 36. Effect of Elevation on Weight If an object weighs m pounds at sea level, then its weight W (in pounds) at a height of h miles above sea level is given approximately by (d) What is the domain of C? What does this domain imply in terms of daily production? (e) Describe the shape of the graph. (f) The point (30, 32 000) is called an inflection point. Describe the behavior of the graph around the inflection point. Cost (dollars per day) (a) Determine the height of the golf ball after it has traveled 100 feet. (b) What is the height after it has traveled 300 feet? (c) What is the height after it has traveled 500 feet? (d) How far was the golf ball hit? (e) Use a graphing utility to graph the function h = h1x2. (f) Use a graphing utility to determine the distance that the ball has traveled when the height of the ball is 90 feet. (g) Create a TABLE with TblStart = 0 and ∆Tbl = 25. To the nearest 25 feet, how far does the ball travel before it reaches a maximum height? What is the maximum height? (h) Adjust the value of ∆Tbl until you determine the distance, to within 1 foot, that the ball travels before it reaches its maximum height. 221 320 160 (5, 30) (15, 90) (0, 30) 0 20 40 Gigabytes 60 g Explaining Concepts: Discussion and Writing 40. Describe how you would find the domain and range of a function if you were given its graph. How would your strategy change if you were given the equation defining the function instead of its graph? 41. How many x-intercepts can the graph of a function have? How many y-intercepts can the graph of a function have? 42. Is a graph that consists of a single point the graph of a function? Can you write the equation of such a function? 222 CHAPTER 3 Functions and Their Graphs 43. Match each of the following functions with the graph that best describes the situation. (a) The cost of building a house as a function of its square footage (b) The height of an egg dropped from a 300-foot building as a function of time (c) The height of a human as a function of time (d) The demand for Big Macs as a function of price (e) The height of a child on a swing as a function of time y y y y x x x x x (III) (II) (I) y (V) (IV) 44. Match each of the following functions with the graph that best describes the situation. (a) The temperature of a bowl of soup as a function of time (b) The number of hours of daylight per day over a 2-year period (c) The population of Florida as a function of time (d) The distance traveled by a car going at a constant velocity as a function of time (e) The height of a golf ball hit with a 7-iron as a function of time y y y x x y x (II) (I) y x 45. Consider the following scenario: Barbara decides to take a walk. She leaves home, walks 2 blocks in 5 minutes at a constant speed, and realizes that she forgot to lock the door. So Barbara runs home in 1 minute. While at her doorstep, it takes her 1 minute to find her keys and lock the door. Barbara walks 5 blocks in 15 minutes and then decides to jog home. It takes her 7 minutes to get home. Draw a graph of Barbara’s distance from home (in blocks) as a function of time. 46. Consider the following scenario: Jayne enjoys riding her bicycle through the woods. At the forest preserve, she gets on her bicycle and rides up a 2000-foot incline in 10 minutes. She then travels down the incline in 3 minutes. The next 5000 feet is level terrain, and she covers the distance in 20 minutes. She rests for 15 minutes. Jayne then travels 10,000 feet in 30 minutes. Draw a graph of Jayne’s distance traveled (in feet) as a function of time. 47. The following sketch represents the distance d (in miles) that Kevin was from home as a function of time t (in hours). Answer the questions by referring to the graph. In parts (a)–(g), how many hours elapsed and how far was Kevin from home during this time? x (V) (IV) (III) (a) (b) (c) (d) (e) (f) (g) (h) (i) From t = 0 to t = 2 From t = 2 to t = 2.5 From t = 2.5 to t = 2.8 From t = 2.8 to t = 3 From t = 3 to t = 3.9 From t = 3.9 to t = 4.2 From t = 4.2 to t = 5.3 What is the farthest distance that Kevin was from home? How many times did Kevin return home? 48. The following sketch represents the speed v (in miles per hour) of Michael’s car as a function of time t (in minutes). v (t ) (7, 50) (2, 30) (8, 38) (4, 30) (4.2, 0) (7.4, 50) (7.6, 38) (6, 0) (9.1, 0) t d (t ) (2, 3) (2.5, 3) (2.8, 0) (3.9, 2.8) (3, 0) (4.2, 2.8) (5.3, 0) t (a) Over what interval of time was Michael traveling fastest? (b) Over what interval(s) of time was Michael’s speed zero? (c) What was Michael’s speed between 0 and 2 minutes? (d) What was Michael’s speed between 4.2 and 6 minutes? (e) What was Michael’s speed between 7 and 7.4 minutes? (f) When was Michael’s speed constant? SECTION 3.3 Properties of Functions 49. Draw the graph of a function whose domain is 5x - 3 … x … 8, x ≠ 56 and whose range is 5y - 1 … y … 2, y ≠ 06. What point(s) in the rectangle - 3 … x … 8, - 1 … y … 2 cannot be on the graph? Compare your graph with those of other students. What differences do you see? 223 50. Is there a function whose graph is symmetric with respect to the x-axis? Explain. 51. Explain why the vertical-line test works. Retain Your Knowledge Problems 52–55 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 52. Factor completely: 8x2 + 24x + 18 53. Find the distance between the points (3, - 6) and (1, 0). 2 54. Write the equation of the line with slope that passes through the point ( - 6, 4). 3 55. Subtract: (4x3 - 5x2 + 2) - (3x2 + 5x - 2) ‘Are You Prepared?’ Answers 1. 1 - 4, 02, 14, 02, 10, - 22, 10, 22 2. False 3.3 Properties of Functions PREPARING FOR THIS SECTION Before getting started, review the following: r Intervals (Section 1.5, pp. 120–121) r Intercepts (Section 2.2, pp. 159–160) r Slope of a Line (Section 2.3, pp. 167–169) r Point–Slope Form of a Line (Section 2.3, p. 171) r Symmetry (Section 2.2, pp. 160–162) Now Work the ‘Are You Prepared?’ problems on page 232. OBJECTIVES 1 Determine Even and Odd Functions from a Graph (p. 223) 2 Identify Even and Odd Functions from an Equation (p. 225) 3 Use a Graph to Determine Where a Function Is Increasing, Decreasing, or Constant (p. 225) 4 Use a Graph to Locate Local Maxima and Local Minima (p. 226) 5 Use a Graph to Locate the Absolute Maximum and the Absolute Minimum (p. 227) 6 Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function Is Increasing or Decreasing (p. 229) 7 Find the Average Rate of Change of a Function (p. 230) To obtain the graph of a function y = f1x2 , it is often helpful to know certain properties that the function has and the impact of these properties on the way the graph will look. 1 Determine Even and Odd Functions from a Graph The words even and odd, when applied to a function f, describe the symmetry that exists for the graph of the function. 224 CHAPTER 3 Functions and Their Graphs A function f is even if and only if, whenever the point 1x, y2 is on the graph of f , the point 1 - x, y2 is also on the graph. Using function notation, we define an even function as follows: DEFINITION A function f is even if, for every number x in its domain, the number - x is also in the domain and f1 - x2 = f1x2 A function f is odd if and only if, whenever the point 1x, y2 is on the graph of f , the point 1 - x, - y2 is also on the graph. Using function notation, we define an odd function as follows: DEFINITION A function f is odd if, for every number x in its domain, the number - x is also in the domain and f1 - x2 = - f1x2 Refer to page 162, where the tests for symmetry are listed. The following results are then evident. THEOREM EX AMPLE 1 A function is even if and only if its graph is symmetric with respect to the y-axis. A function is odd if and only if its graph is symmetric with respect to the origin. Determining Even and Odd Functions from the Graph Determine whether each graph given in Figure 17 is the graph of an even function, an odd function, or a function that is neither even nor odd. y y x Figure 17 Solution y x (b) (a) x (c) (a) The graph in Figure 17(a) is that of an even function, because the graph is symmetric with respect to the y-axis. (b) The function whose graph is given in Figure 17(b) is neither even nor odd, because the graph is neither symmetric with respect to the y-axis nor symmetric with respect to the origin. (c) The function whose graph is given in Figure 17(c) is odd, because its graph is symmetric with respect to the origin. r Now Work PROBLEMS 25(a), (b), AND (d) SECTION 3.3 Properties of Functions 225 2 Identify Even and Odd Functions from an Equation EXAMPL E 2 Identifying Even and Odd Functions Algebraically Determine whether each of the following functions is even, odd, or neither. Then determine whether the graph is symmetric with respect to the y-axis, with respect to the origin, or neither. (a) f1x2 = x2 - 5 (c) h 1x2 = 5x3 - x Solution (b) g1x2 = x3 - 1 (d) F1x2 = 0 x 0 (a) To determine whether f is even, odd, or neither, replace x by - x in f1x2 = x2 - 5. f 1 - x2 = 1 - x2 2 - 5 = x2 - 5 = f1x2 Since f 1 - x2 = f1x2 , the function is even, and the graph of f is symmetric with respect to the y-axis. (b) Replace x by - x in g1x2 = x3 - 1. g1 - x2 = 1 - x2 3 - 1 = - x3 - 1 Since g1 - x2 ≠ g1x2 and g1- x2 ≠ - g1x2 = - 1x3 - 12 = - x3 + 1, the function is neither even nor odd. The graph of g is not symmetric with respect to the y-axis, nor is it symmetric with respect to the origin. (c) Replace x by - x in h 1x2 = 5x3 - x. h 1 - x2 = 51 - x2 3 - 1 - x2 = - 5x3 + x = - 15x3 - x2 = - h 1x2 Since h 1 - x2 = - h 1x2, h is an odd function, and the graph of h is symmetric with respect to the origin. (d) Replace x by - x in F1x2 = 0 x 0 . F1 - x2 = 0 - x 0 = 0 - 1 0 # 0 x 0 = 0 x 0 = F1x2 Since F1 - x2 = F1x2, F is an even function, and the graph of F is symmetric with respect to the y-axis. r Now Work PROBLEM 37 3 Use a Graph to Determine Where a Function Is Increasing, Decreasing, or Constant Consider the graph given in Figure 18. If you look from left to right along the graph of the function, you will notice that parts of the graph are going up, parts are going down, and parts are horizontal. In such cases, the function is described as increasing, decreasing, or constant, respectively. y 5 (0, 4) (6, 0) (2, 0) 4 Figure 18 EXAMPL E 3 (4, 2) (3, 4) y = f (x ) (6, 1) 6 x 2 Determining Where a Function Is Increasing, Decreasing, or Constant from Its Graph Determine the values of x for which the function in Figure 18 is increasing. Where is it decreasing? Where is it constant? 226 CHAPTER 3 Functions and Their Graphs Solution When determining where a function is increasing, where it is decreasing, and where it is constant, we use strict inequalities involving the independent variable x, or we use open intervals* of x-coordinates. The function whose graph is given in Figure 18 is increasing on the open interval 1 - 4, 02 , or for - 4 6 x 6 0. The function is decreasing on the open intervals 1 - 6, - 42 and 13, 62 , or for - 6 6 x 6 - 4 and 3 6 x 6 6. The function is constant on the open interval 10, 32 , or for 0 6 x 6 3. WARNING Describe the behavior of a graph in terms of its x-values. Do not say the graph in Figure 18 is increasing from the point 1 - 4, - 22 to the point (0, 4). Rather, say it is increasing on the interval ■ 1 - 4, 02 . DEFINITIONS A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 6 f1x2 2. In Words If a function is decreasing, then as the values of x get bigger, the values of the function get smaller. If a function is increasing, then as the values of x get bigger, the values of the function also get bigger. If a function is constant, then as the values of x get bigger, the values of the function remain unchanged. A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 7 f 1x2 2. A function f is constant on an open interval I if, for all choices of x in I, the values f1x2 are equal. Figure 19 illustrates the definitions. The graph of an increasing function goes up from left to right, the graph of a decreasing function goes down from left to right, and the graph of a constant function remains at a fixed height. y y y f (x 1) f (x 2) f (x 1) x1 x2 x f (x 1) f (x 2) x1 x x2 x1 (b) For x 1 < x 2 in l, f(x 1) > f (x 2); f is decreasing on I. (a) For x 1 < x 2 in l, f (x 1) < f (x 2); f is increasing on I. Now Work PROBLEMS f(x 2) x2 x l l l Figure 19 r More precise definitions follow: (c) For all x in I, the values of f are equal; f is constant on I. 13, 15, 17, AND 25(c) 4 Use a Graph to Locate Local Maxima and Local Minima Suppose f is a function defined on an open interval I containing c. If the value of f at c is greater than or equal to the values of f on I, then f has a local maximum at c.† See Figure 20(a). If the value of f at c is less than or equal to the values of f on I, then f has a local minimum at c. See Figure 20(b). y y f has a local maximum f(c) at c. (c, f(c)) c (a) f(c) x f has a local minimum at c. (c, f(c)) c (b) Figure 20 Local maximum and local minimum *The open interval 1a, b2 consists of all real numbers x for which a 6 x 6 b. † Some texts use the term relative instead of local. x SECTION 3.3 Properties of Functions DEFINITIONS 227 Let f be a function defined on some interval I. A function f has a local maximum at c if there is an open interval in I containing c so that, for all x in this open interval, we have f1x2 … f1c2. We call f1c2 a local maximum value of f. A function f has a local minimum at c if there is an open interval in I containing c so that, for all x in this open interval, we have f 1x2 Ú f1c2. We call f1c2 a local minimum value of f. If f has a local maximum at c, then the value of f at c is greater than or equal to the values of f near c. If f has a local minimum at c, then the value of f at c is less than or equal to the values of f near c. The word local is used to suggest that it is only near c, not necessarily over the entire domain, that the value f1c2 has these properties. EXAMPL E 4 y y f(x) Figure 21 shows the graph of a function f. (1, 2) 2 (–1, 1) –2 3 Finding Local Maxima and Local Minima from the Graph of a Function and Determining Where the Function Is Increasing, Decreasing, or Constant x Figure 21 Solution WARNING The y-value is the local maximum value or local minimum value, and it occurs at some x-value. For example, in Figure 21, we say f has a local maximum at 1 and the local maximum value is 2. ■ (a) At what value(s) of x, if any, does f have a local maximum? List the local maximum values. (b) At what value(s) of x, if any, does f have a local minimum? List the local minimum values. (c) Find the intervals on which f is increasing. Find the intervals on which f is decreasing. The domain of f is the set of real numbers. (a) f has a local maximum at 1, since for all x close to 1, we have f 1x2 … f112 . The local maximum value is f112 = 2. (b) f has local minima at - 1 and at 3. The local minimum values are f1 - 12 = 1 and f132 = 0. (c) The function whose graph is given in Figure 21 is increasing for all values of x between - 1 and 1 and for all values of x greater than 3. That is, the function is increasing on the intervals 1 - 1, 12 and 13, q 2 , or for - 1 6 x 6 1 and x 7 3. The function is decreasing for all values of x less than - 1 and for all values of x between 1 and 3. That is, the function is decreasing on the intervals 1 - q , - 12 and 11, 32 , or for x 6 - 1 and 1 6 x 6 3. r Now Work y y f (x) (b, f (b)) (v, f (v)) a u v b domain: [a, b] for all x in [a, b], f(x) f(u) for all x in [a, b], f(x) f(v) absolute maximum: f(u) absolute minimum: f(v) Figure 22 19 AND 21 5 Use a Graph to Locate the Absolute Maximum and the Absolute Minimum (u, f(u)) (a, f(a)) PROBLEMS Look at the graph of the function f given in Figure 22. The domain of f is the closed interval 3 a, b4 . Also, the largest value of f is f 1u2 and the smallest value of f is f1v2. These are called, respectively, the absolute maximum and the absolute minimum of f on 3 a, b 4 . x DEFINITION Let f be a function defined on some interval I. If there is a number u in I for which f 1x2 … f1u2 for all x in I, then f has an absolute maximum at u, and the number f1u2 is the absolute maximum of f on I. If there is a number v in I for which f1x2 Ú f 1v2 for all x in I, then f has an absolute minimum at v, and the number f1v2 is the absolute minimum of f on I. 228 CHAPTER 3 Functions and Their Graphs The absolute maximum and absolute minimum of a function f are sometimes called the extreme values of f on I. The absolute maximum or absolute minimum of a function f may not exist. Let’s look at some examples. Finding the Absolute Maximum and the Absolute Minimum from the Graph of a Function EX AMPLE 5 For each graph of a function y = f 1x2 in Figure 23, find the absolute maximum and the absolute minimum, if they exist. Also, find any local maxima or local minima. (3, 6) 6 4 6 6 (5, 5) 4 (4, 4) 2 2 1 3 5 x 1 Figure 23 3 (b) Solution WARNING A function may have an absolute maximum or an absolute minimum at an endpoint but not a local maximum or a local minimum. Why? Local maxima and local minima are found over some open interval I, and this interval cannot be created around an endpoint. ■ 4 4 (1, 4) x 2 2 (1, 1) (2, 1) 5 6 (4, 3) 2 (3, 1) (a) 6 (0, 3) (1, 2) (0, 1) (5, 4) 4 (5, 3) y y y y y 1 3 (c) (2, 2) (0, 0) 5 x 1 3 (d) 5 x 1 3 5 (e) (a) The function f whose graph is given in Figure 23(a) has the closed interval [0, 5] as its domain. The largest value of f is f132 = 6, the absolute maximum. The smallest value of f is f102 = 1, the absolute minimum. The function has a local maximum of 6 at x = 3 and a local minimum of 4 at x = 4. (b) The function f whose graph is given in Figure 23(b) has the domain 5 x|1 … x … 5, x ≠ 36 . Note that we exclude 3 from the domain because of the “hole” at (3, 1). The largest value of f on its domain is f 152 = 3, the absolute maximum. There is no absolute minimum. Do you see why? As you trace the graph, getting closer to the point (3, 1), there is no single smallest value. [As soon as you claim a smallest value, we can trace closer to (3, 1) and get a smaller value!] The function has no local maxima or minima. (c) The function f whose graph is given in Figure 23(c) has the interval [0, 5] as its domain. The absolute maximum of f is f152 = 4. The absolute minimum is 1. Notice that the absolute minimum 1 occurs at any number in the interval [1, 2]. The function has a local minimum value of 1 at every x in the interval [1, 2], but it has no local maximum value. (d) The function f given in Figure 23(d) has the interval 3 0, q ) as its domain. The function has no absolute maximum; the absolute minimum is f102 = 0. The function has no local maximum or local minimum. (e) The function f in Figure 23(e) has the domain 5 x|1 6 x 6 5, x ≠ 26. The function has no absolute maximum and no absolute minimum. Do you see why? The function has a local maximum value of 3 at x = 4, but no local minimum value. r In calculus, there is a theorem with conditions that guarantee a function will have an absolute maximum and an absolute minimum. THEOREM Extreme Value Theorem If f is a continuous function* whose domain is a closed interval 3 a, b 4 , then f has an absolute maximum and an absolute minimum on 3 a, b4 . *Although a precise definition requires calculus, we’ll agree for now that a continuous function is one whose graph has no gaps or holes and can be traced without lifting the pencil from the paper. x SECTION 3.3 Properties of Functions 229 The absolute maximum (minimum) can be found by selecting the largest (smallest) value of f from the following list: 1. The values of f at any local maxima or local minima of f in [a, b]. 2. The value of f at each endpoint of [a, b]—that is, f(a) and f(b). For example, the graph of the function f given in Figure 23(a) is continuous on the closed interval [0, 5]. The Extreme Value Theorem guarantees that f has extreme values on [0, 5]. To find them, we list 1. The value of f at the local extrema: f 132 = 6, f142 = 4 2. The value of f at the endpoints: f 102 = 1, f152 = 5 The largest of these, 6, is the absolute maximum; the smallest of these, 1, is the absolute minimum. Now Work PROBLEM 49 6 Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function Is Increasing or Decreasing To locate the exact value at which a function f has a local maximum or a local minimum usually requires calculus. However, a graphing utility may be used to approximate these values using the MAXIMUM and MINIMUM features. EXAM PL E 6 Using a Graphing Utility to Approximate Local Maxima and Minima and to Determine Where a Function Is Increasing or Decreasing (a) Use a graphing utility to graph f1x2 = 6x3 - 12x + 5 for - 2 6 x 6 2. Approximate where f has a local maximum and where f has a local minimum. (b) Determine where f is increasing and where it is decreasing. Solution (a) Graphing utilities have a feature that finds the maximum or minimum point of a graph within a given interval. Graph the function f for - 2 6 x 6 2. The MAXIMUM and MINIMUM commands require us to first determine the open interval I. The graphing utility will then approximate the maximum or minimum value in the interval. Using MAXIMUM, we find that the local maximum value is 11.53 and that it occurs at x = - 0.82, rounded to two decimal places. See Figure 24(a). Using MINIMUM, we find that the local minimum value is - 1.53 and that it occurs at x = 0.82, rounded to two decimal places. See Figure 24(b). 30 30 −2 −2 2 −10 −10 Figure 24 (a) Local maximum 2 (b) Local minimum (b) Looking at Figures 24(a) and (b), we see that the graph of f is increasing from x = - 2 to x = - 0.82 and from x = 0.82 to x = 2, so f is increasing on the intervals 1 - 2, - 0.822 and 10.82, 22 , or for - 2 6 x 6 - 0.82 and 0.82 6 x 6 2. The graph is decreasing from x = - 0.82 to x = 0.82, so f is decreasing on the interval 1 - 0.82, 0.822 , or for - 0.82 6 x 6 0.82. r Now Work PROBLEM 57 230 CHAPTER 3 Functions and Their Graphs 7 Find the Average Rate of Change of a Function In Section 2.3, we said that the slope of a line can be interpreted as the average rate of change. To find the average rate of change of a function between any two points on its graph, calculate the slope of the line containing the two points. If a and b, a ≠ b, are in the domain of a function y = f 1x2, the average rate of change of f from a to b is defined as DEFINITION In Words Average rate of change = The symbol ∆ is the Greek capital letter delta and is read “change in.” f1b2 - f1a2 ∆y = ∆x b - a a ≠ b (1) The symbol ∆y in equation (1) is the “change in y,” and ∆x is the “change in x.” The average rate of change of f is the change in y divided by the change in x. Finding the Average Rate of Change EX AMPLE 7 Find the average rate of change of f1x2 = 3x2: (a) From 1 to 3 (b) From 1 to 5 (c) From 1 to 7 (a) The average rate of change of f1x2 = 3x2 from 1 to 3 is Solution f132 - f112 ∆y 27 - 3 24 = = = = 12 ∆x 3 - 1 3 - 1 2 (b) The average rate of change of f 1x2 = 3x2 from 1 to 5 is y f152 - f112 ∆y 75 - 3 72 = = = = 18 ∆x 5 - 1 5 - 1 4 160 (7, 147) 120 Average rate of change 5 24 80 (c) The average rate of change of f 1x2 = 3x2 from 1 to 7 is (5, 75) Average rate of change 5 18 40 (3, 27) (1, 3) (0, 0) 2 4 Figure 25 f 1x2 = 3x2 Average rate of change 5 12 6 x f172 - f112 ∆y 147 - 3 144 = = = = 24 ∆x 7 - 1 7 - 1 6 r See Figure 25 for a graph of f 1x2 = 3x2. The function f is increasing for x 7 0. The fact that the average rate of change is positive for any x1, x2, x1 ≠ x2, in the interval 11, 72 indicates that the graph is increasing on 1 6 x 6 7. Further, the average rate of change is consistently getting larger for 1 6 x 6 7, which indicates that the graph is increasing at an increasing rate. Now Work PROBLEM 65 The Secant Line The average rate of change of a function has an important geometric interpretation. Look at the graph of y = f 1x2 in Figure 26. Two points are labeled on the graph: 1a, f1a2 2 and 1b, f1b2 2. The line containing these two points is called the secant line; its slope is msec = f 1b2 - f 1a2 b - a SECTION 3.3 Properties of Functions y 231 y 5 f (x ) Secant line (b, f (b )) I(b ) 2 f (a ) (a, f (a )) b2a b a x Figure 26 Secant line THEOREM Slope of the Secant Line The average rate of change of a function from a to b equals the slope of the secant line containing the two points 1 a, f 1a2 2 and 1b, f 1b2 2 on its graph. EXAMPL E 8 Finding the Equation of a Secant Line Suppose that g1x2 = 3x2 - 2x + 3. (a) Find the average rate of change of g from - 2 to 1. (b) Find an equation of the secant line containing 1 - 2, g1 - 22 2 and 11, g112 2 . (c) Using a graphing utility, draw the graph of g and the secant line obtained in part (b) on the same screen. Solution (a) The average rate of change of g1x2 = 3x2 - 2x + 3 from - 2 to 1 is Average rate of change = g112 - g1 - 22 1 - 1 - 22 4 - 19 3 15 = = -5 3 = g (1) = 3(1)2 - 2(1) + 3 = 4 g ( - 2) = 3( - 2)2 - 2( - 2) + 3 = 19 (b) The slope of the secant line containing 1 - 2, g1 - 22 2 = 1 - 2, 192 and 11, g112 2 = 11, 42 is msec = - 5. Use the point–slope form to find an equation of the secant line. 24 −3 y - y1 = msec 1x - x1 2 3 −4 Figure 27 Graph of g and the secant line Point–slope form of the secant line y - 19 = - 51x - 1 - 22 2 x1 = - 2, y1 = g ( - 2) = 19, msec = - 5 y - 19 = - 5x - 10 Distribute. y = - 5x + 9 Slope–intercept form of the secant line (c) Figure 27 shows the graph of g along with the secant line y = - 5x + 9. Now Work PROBLEM 71 r 232 CHAPTER 3 Functions and Their Graphs 3.3 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. The interval 12, 52 can be written as the inequality (pp. 120–121) . 2. The slope of the line containing the points 1 - 2, 32 and 13, 82 is . (pp. 167–169) 4. Write the point–slope form of the line with slope 5 containing the point 13, - 22. (p. 171) 5. The intercepts of the equation y = x2 - 9 are (pp. 159–160) . 3. Test the equation y = 5x2 - 1 for symmetry with respect to the x-axis, the y-axis, and the origin. (pp. 160–162) Concepts and Vocabulary 6. A function f is on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2. function f is one for which f 1 - x2 = f 1x2 for 7. A(n) every x in the domain of f; a(n) function f is one for which f 1 - x2 = - f 1x2 for every x in the domain of f. 8. True or False A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 7 f 1x2 2. 9. True or False A function f has a local maximum at c if there is an open interval I containing c such that for all x in I, f 1x2 … f 1c2. 10. True or False Even functions have graphs that are symmetric with respect to the origin. 11. An odd function is symmetric with respect to (a) the x-axis (b) the y-axis (c) the origin (d) the line y = x . 12. Which of the following intervals is required to guarantee a continuous function will have both an absolute maximum and an absolute minimum? (a) (a, b) (b) (a, b4 (d) 3a, b4 (c) 3a, b) Skill Building In Problems 13–24, use the graph of the function f given. y 13. Is f increasing on the interval 1 - 8, - 22? (2, 10) 10 14. Is f decreasing on the interval 1 - 8, - 42? (22, 6) 15. Is f increasing on the interval 1 - 2, 62? (210, 0) 16. Is f decreasing on the interval 12, 52? (5, 0) (25, 0) 210 17. List the interval(s) on which f is increasing. 18. List the interval(s) on which f is decreasing. 25 (0, 0) (28, 24) 19. Is there a local maximum at 2? If yes, what is it? (7, 3) 10 x 5 26 20. Is there a local maximum at 5? If yes, what is it? 21. List the number(s) at which f has a local maximum. What are the local maximum values? 22. List the number(s) at which f has a local minimum. What are the local minimum values? 23. Find the absolute minimum of f on 3 - 10, 74 . 24. Find the absolute maximum of f on 3 - 10, 74 . In Problems 25–32, the graph of a function is given. Use the graph to find: (a) The intercepts, if any (b) The domain and range (c) The intervals on which the function is increasing, decreasing, or constant (d) Whether the function is even, odd, or neither 25. 26. y 4 (4, 2) (3, 3) 27. y 3 (0, 3) 28. y 3 (3, 3) y 3 (0, 2) (4, 2) (0, 1) 4 (2, 0) (2, 0) 4x 3 (1, 0) (1, 0) 3 x 3 3 x 3 (1, 0) 3x 233 SECTION 3.3 Properties of Functions 29. 30. y 2 31. y 2 (––2 , 1) ( – 3, 2) (0, 1) 32. y 3 1 0, – 2 ( ) (3, 1) ( –1, 2) –– 2 –– 2 x 2 (, 1) ( ––2 , 1) 2 x 1 –, 0 3 ( ) 2 2 (2, 2) (2, 1) (2.3, 0) (3, 0) (0, 1) 3 3 x (2, –1) (1, –1) –3 (, 1) y 3 3 x (3, 2) –3 2 In Problems 33–36, the graph of a function f is given. Use the graph to find: (a) The numbers, if any, at which f has a local maximum. What are the local maximum values? (b) The numbers, if any, at which f has a local minimum. What are the local minimum values? 33. 34. y 4 35. y 36. y ( 3 (0, 3) 1 (0, 2) 4 (2, 0) 4x (2, 0) 3 (1, 0) 3 x (1, 0) –– 2 –– , 2 1) –– 2 π x π2 π 2 π x (π, 1) 1 ( ––2 , 1) y 2 (π, 1) 2 In Problems 37–48, determine algebraically whether each function is even, odd, or neither. 37. f 1x2 = 4x3 38. f 1x2 = 2x4 - x2 39. g1x2 = - 3x2 - 5 40. h1x2 = 3x3 + 5 3 41. F 1x2 = 2 x 42. G1x2 = 1x 43. f 1x2 = x + 0 x 0 3 44. f 1x2 = 2 2x2 + 1 45. g1x2 = 1 x2 46. h1x2 = x x - 1 47. h1x2 = 2 - x3 3x2 - 9 48. F 1x2 = 2x 0x0 In Problems 49–56, for each graph of a function y = f(x), find the absolute maximum and the absolute minimum, if they exist. Identify any local maximum values or local minimum values. 49. 50. y (1, 4) 4 (4, 4) 4 (3, 3) 2 51. y (5, 1) 53. (3, 4) 4 (2, 4) 4 2 (1, 1) x 5 1 54. y 4 2 (0, 1) (1, 1) (5, 0) 3 5 x 1 55. y (2, 4) 4 (1, 3) (4, 3) (0, 2) 2 3 y (0, 3) (2, 2) 1 52. y 3 5 x 1 56. y (1, 3) (2, 3) 2 (2, 0) 2 (4, 1) 2 (3, 2) (0, 2) 1 (0, 2) (3, 1) x 3 1 (2, 0) 3 (0, 0) 1 x y (3, 2) 2 3 x 3 1 3 x x In Problems 57–64, use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. 57. f 1x2 = x3 - 3x + 2 59. f 1x2 = x5 - x3 1 - 2, 22 58. f 1x2 = x3 - 3x2 + 5 60. f 1x2 = x4 - x2 1 - 2, 22 61. f 1x2 = - 0.2x3 - 0.6x2 + 4x - 6 63. f 1x2 = 0.25x4 + 0.3x3 - 0.9x2 + 3 1 - 6, 42 1 - 3, 22 1 - 1, 32 1 - 2, 22 62. f 1x2 = - 0.4x3 + 0.6x2 + 3x - 2 1 - 4, 52 64. f 1x2 = - 0.4x4 - 0.5x3 + 0.8x2 - 2 1 - 3, 22 234 CHAPTER 3 Functions and Their Graphs 65. Find the average rate of change of f 1x2 = - 2x2 + 4: (a) From 0 to 2 (b) From 1 to 3 (c) From 1 to 4 66. Find the average rate of change of f 1x2 = - x3 + 1: (a) From 0 to 2 (b) From 1 to 3 (c) From - 1 to 1 70. f 1x2 = - 4x + 1 (a) Find the average rate of change from 2 to 5. (b) Find an equation of the secant line containing 12, f 122 2 and 15, f 152 2. 71. g1x2 = x2 - 2 (a) Find the average rate of change from - 2 to 1. (b) Find an equation of the secant line containing 1 - 2, g1 - 22 2 and 11, g112 2. 67. Find the average rate of change of g1x2 = x3 - 2x + 1: (a) From - 3 to - 2 (b) From - 1 to 1 (c) From 1 to 3 72. g1x2 = x2 + 1 (a) Find the average rate of change from - 1 to 2. (b) Find an equation of the secant line containing 1 - 1, g1 - 12 2 and 12, g122 2. 68. Find the average rate of change of h1x2 = x2 - 2x + 3: (a) From - 1 to 1 (b) From 0 to 2 (c) From 2 to 5 73. h1x2 = x2 - 2x (a) Find the average rate of change from 2 to 4. (b) Find an equation of the secant line containing 12, h122 2 and 14, h142 2. 69. f 1x2 = 5x - 2 (a) Find the average rate of change from 1 to 3. (b) Find an equation of the secant line containing 11, f 112 2 and 13, f 132 2. 74. h1x2 = - 2x2 + x (a) Find the average rate of change from 0 to 3. (b) Find an equation of the secant line containing 10, h102 2 and 13, h132 2. Mixed Practice 75. g(x) = x3 - 27x (a) Determine whether g is even, odd, or neither. (b) There is a local minimum value of - 54 at 3. Determine the local maximum value. 76. f(x) = - x3 + 12x (a) Determine whether f is even, odd, or neither. (b) There is a local maximum value of 16 at 2. Determine the local minimum value. 77. F(x) = - x4 + 8x2 + 8 (a) Determine whether F is even, odd, or neither. (b) There is a local maximum value of 24 at x = 2. Determine a second local maximum value. (c) Suppose the area under the graph of F between x = 0 and x = 3 that is bounded from below by the x-axis is 47.4 square units. Using the result from part (a), determine the area under the graph of F between x = - 3 and x = 0 that is bounded from below by the x-axis. 78. G(x) = - x4 + 32x2 + 144 (a) Determine whether G is even, odd, or neither. (b) There is a local maximum value of 400 at x = 4. Determine a second local maximum value. (c) Suppose the area under the graph of G between x = 0 and x = 6 that is bounded from below by the x-axis is 1612.8 square units. Using the result from part (a), determine the area under the graph of G between x = - 6 and x = 0 that is bounded from below by the x-axis. Applications and Extensions 79. Minimum Average Cost The average cost per hour in dollars, C, of producing x riding lawn mowers can be modeled by the function C 1x2 = 0.3x2 + 21x - 251 + 2500 x (a) Use a graphing utility to graph C = C 1x2 . (b) Determine the number of riding lawn mowers to produce in order to minimize average cost. (c) What is the minimum average cost? 80. Medicine Concentration The concentration C of a medication in the bloodstream t hours after being administered is modeled by the function C 1t2 = - 0.002x4 + 0.039t 3 - 0.285t 2 + 0.766t + 0.085 (a) After how many hours will the concentration be highest? (b) A woman nursing a child must wait until the concentration is below 0.5 before she can feed him. After taking the medication, how long must she wait before feeding her child? 81. Data Plan Cost The monthly cost C, in dollars, for wireless data plans with x gigabytes of data included is shown in the table below. Since each input value for x corresponds to exactly one output value for C, the plan cost is a function of the number of data gigabytes. Thus C(x) represents the monthly cost for a wireless data plan with x gigabytes included. GB Cost ($) GB Cost ($) 4 70 20 150 6 80 30 225 10 100 40 300 15 130 50 375 (a) Plot the points (4, 70), (6, 80), (10, 100), and so on in a Cartesian plane. (b) Draw a line segment from the point (10, 100) to (30, 225). What does the slope of this line segment represent? SECTION 3.3 Properties of Functions (c) Find the average rate of change of the monthly cost from 4 to 10 gigabytes. (d) Find the average rate of change of the monthly cost from 10 to 30 gigabytes. (e) Find the average rate of change of the monthly cost from 30 to 50 gigabytes. (f) What is happening to the average rate of change as the gigabytes of data increase? 82. National Debt The size of the total debt owed by the United States federal government continues to grow. In fact, according to the Department of the Treasury, the debt per person living in the United States is approximately $53,000 (or over $140,000 per U.S. household). The following data represent the U.S. debt for the years 2001–2013. Since the debt D depends on the year y, and each input corresponds to exactly one output, the debt is a function of the year. So D(y) represents the debt for each year y. Year Debt (billions of dollars) Year Debt (billions of dollars) 2001 5807 2008 10,025 2002 6228 2009 11,910 2003 6783 2010 13,562 2004 7379 2011 14,790 2005 7933 2012 16,066 2006 8507 2013 16,738 2007 9008 (a) Find the average rate of change of the population from 0 to 2.5 hours. (b) Find the average rate of change of the population from 4.5 to 6 hours. (c) What is happening to the average rate of change as time passes? 84. e-Filing Tax Returns The Internal Revenue Service Restructuring and Reform Act (RRA) was signed into law by President Bill Clinton in 1998. A major objective of the RRA was to promote electronic filing of tax returns. The data in the table that follows show the percentage of individual income tax returns filed electronically for filing years 2004–2012. Since the percentage P of returns filed electronically depends on the filing year y, and each input corresponds to exactly one output, the percentage of returns filed electronically is a function of the filing year; so P 1y2 represents the percentage of returns filed electronically for filing year y. (a) Find the average rate of change of the percentage of e-filed returns from 2004 to 2006. (b) Find the average rate of change of the percentage of e-filed returns from 2007 to 2009. (c) Find the average rate of change of the percentage of e-filed returns from 2010 to 2012. (d) What is happening to the average rate of change as time passes? Source: www.treasurydirect.gov (a) Plot the points (2001, 5807), (2002, 6228), and so on in a Cartesian plane. (b) Draw a line segment from the point (2001, 5807) to (2006, 8507). What does the slope of this line segment represent? (c) Find the average rate of change of the debt from 2002 to 2004. (d) Find the average rate of change of the debt from 2006 to 2008. (e) Find the average rate of change of the debt from 2010 to 2012. (f) What appears to be happening to the average rate of change as time passes? 83. E. coli Growth A strain of E. coli Beu 397-recA441 is placed into a nutrient broth at 30° Celsius and allowed to grow. The data shown in the table are collected. The population is measured in grams and the time in hours. Since population P depends on time t, and each input corresponds to exactly one output, we can say that population is a function of time. Thus P 1t2 represents the population at time t. Time (hours), t Population (grams), P 0 0.09 2.5 0.18 3.5 0.26 4.5 0.35 6 0.50 235 Year Percentage of returns e-filed 2004 46.5 2005 51.1 2006 53.8 2007 57.1 2008 58.5 2009 67.2 2010 69.8 2011 77.2 2012 82.7 Source: Internal Revenue Service 85. For the function f 1x2 = x2, compute the average rate of change: (a) From 0 to 1 (b) From 0 to 0.5 (c) From 0 to 0.1 (d) From 0 to 0.01 (e) From 0 to 0.001 (f) Use a graphing utility to graph each of the secant lines along with f . (g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? 236 CHAPTER 3 Functions and Their Graphs 86. For the function f 1x2 = x2, compute the average rate of change: (a) From 1 to 2 (b) From 1 to 1.5 (c) From 1 to 1.1 (d) From 1 to 1.01 (e) From 1 to 1.001 (f) Use a graphing utility to graph each of the secant lines along with f . (g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? Problems 87–94 require the following discussion of a secant line. The slope of the secant line containing the two points 1x, f 1x2 2 and 1x + h, f 1x + h2 2 on the graph of a function y = f 1x2 may be given as msec = f 1x + h2 - f 1x2 1x + h2 - x = f 1x + h2 - f 1x2 h , h ≠ 0 In calculus, this expression is called the difference quotient of f. (a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer. (b) Find msec for h = 0.5, 0.1, and 0.01 at x = 1. What value does msec approach as h approaches 0? (c) Find an equation for the secant line at x = 1 with h = 0.01. (d) Use a graphing utility to graph f and the secant line found in part (c) in the same viewing window. 87. f 1x2 = 2x + 5 88. f 1x2 = - 3x + 2 89. f 1x2 = x2 + 2x 90. f 1x2 = 2x2 + x 91. f 1x2 = 2x2 - 3x + 1 92. f 1x2 = - x2 + 3x - 2 93 f 1x2 = 94. f 1x2 = 1 x 1 x2 Explaining Concepts: Discussion and Writing 95. Draw the graph of a function that has the following properties: domain: all real numbers; range: all real numbers; intercepts: 10, - 32 and 13, 02; a local maximum value of - 2 is at - 1; a local minimum value of - 6 is at 2. Compare your graph with those of others. Comment on any differences. 96. Redo Problem 95 with the following additional information: increasing on 1 - q , - 12, 12, q 2; decreasing on 1 - 1, 22. Again compare your graph with others and comment on any differences. 97. How many x-intercepts can a function defined on an interval have if it is increasing on that interval? Explain. 99. Can a function be both even and odd? Explain. 100. Using a graphing utility, graph y = 5 on the interval 1 - 3, 32. Use MAXIMUM to find the local maximum values on 1 - 3, 32. Comment on the result provided by the calculator. 101. A function f has a positive average rate of change on the interval 3 2, 5 4 . Is f increasing on 3 2, 5 4 ? Explain. 102. Show that a constant function f(x) = b has an average rate of change of 0. Compute the average rate of change of y = 24 - x2 on the interval 3 - 2, 24. Explain how this can happen. 98. Suppose that a friend of yours does not understand the idea of increasing and decreasing functions. Provide an explanation, complete with graphs, that clarifies the idea. Retain Your Knowledge Problems 103–106 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 103. Write each number in scientific notation. (a) 0.00000701 (b) 2, 305, 000, 000 104. Simplify: 2540 105. Solve: 14 6 5 - 3x … 29 106. The shelf life of a perishable commodity varies inversely with the storage temperature. If the shelf life at 10°C is 33 days, what is the shelf life at 40°C? ‘Are You Prepared?’ Answers 1. 2 6 x 6 5 2. 1 3. symmetric with respect to the y-axis 4. y + 2 = 51x - 32 5. 1 - 3, 02, 13, 02, 10, - 92 SECTION 3.4 Library of Functions; Piecewise-defined Functions 237 3.4 Library of Functions; Piecewise-defined Functions PREPARING FOR THIS SECTION Before getting started, review the following: r Intercepts (Section 2.2, pp. 159–160) r Graphs of Key Equations (Section 2.2: Example 3, p. 159; Example 10, p. 163; Example 11, p. 163; Example 12, p. 164) Now Work the ‘Are You Prepared?’ problems on page 244. OBJECTIVES 1 Graph the Functions Listed in the Library of Functions (p. 237) 2 Graph Piecewise-defined Functions (p. 242) 1 Graph the Functions Listed in the Library of Functions First we introduce a few more functions, beginning with the square root function. On page 163, we graphed the equation y = 1x. Figure 28 shows a graph of the function f1x2 = 1x. Based on the graph, we have the following properties: Properties of f (x) = !x y 6 (1, 1) (4, 2) (9, 3) (0, 0) –2 5 10 x Figure 28 Square root function EXAMPL E 1 1. The domain and the range are the set of nonnegative real numbers. 2. The x-intercept of the graph of f1x2 = 1x is 0. The y-intercept of the graph of f1x2 = 1x is also 0. 3. The function is neither even nor odd. 4. The function is increasing on the interval 10, q 2. 5. The function has an absolute minimum of 0 at x = 0. Graphing the Cube Root Function 3 (a) Determine whether f1x2 = 2 x is even, odd, or neither. State whether the graph of f is symmetric with respect to the y-axis or symmetric with respect to the origin. 3 (b) Determine the intercepts, if any, of the graph of f1x2 = 2 x. 3 (c) Graph f1x2 = 2 x. Solution (a) Because 3 3 f 1 - x2 = 2 -x = - 2 x = - f1x2 the function is odd. The graph of f is symmetric with respect to the origin. 3 (b) The y-intercept is f102 = 2 0 = 0. The x-intercept is found by solving the equation f 1x2 = 0. f1x2 = 0 3 2 x = 0 x = 0 3 f (x) = 2 x Cube both sides of the equation. The x-intercept is also 0. (c) Use the function to form Table 4 (on page 238) and obtain some points on the graph. Because of the symmetry with respect to the origin, we find only points 3 1x, y2 for which x Ú 0. Figure 29 shows the graph of f1x2 = 2x. 238 CHAPTER 3 Functions and Their Graphs Table 4 x y = f (x) = !x 0 0 (0, 0) 1 8 1 2 1 1 a , b 8 2 1 1 (1, 1) 3 2 3 ( 1–8 , 1–2 2 (1, 1) ) (2, 2 ) ( 1–8 , 1–2) 3 3 1 2, 2 22 3 2 2 ≈ 1.26 8 y 3 (x, y) 3 x (0, 0) 3 (1, 1) (2, 2 ) (8, 2) 3 Figure 29 Cube Root Function r From the results of Example 1 and Figure 29, we have the following properties of the cube root function. Properties of f(x) = !x 3 1. The domain and the range are the set of all real numbers. 3 2. The x-intercept of the graph of f1x2 = 2 x is 0. The y-intercept of the 3 graph of f1x2 = 2 x is also 0. 3. The function is odd. The graph is symmetric with respect to the origin. 4. The function is increasing on the interval 1 - q , q 2. 5. The function does not have any local minima or any local maxima. EX AMPLE 2 Solution Graphing the Absolute Value Function (a) Determine whether f 1x2 = 0 x 0 is even, odd, or neither. State whether the graph of f is symmetric with respect to the y-axis, symmetric with respect to the origin, or neither. (b) Determine the intercepts, if any, of the graph of f1x2 = 0 x 0 . (c) Graph f1x2 = 0 x 0 . (a) Because f1 - x2 = 0 - x 0 = 0 x 0 = f1x2 the function is even. The graph of f is symmetric with respect to the y-axis. (b) The y-intercept is f102 = 0 0 0 = 0. The x-intercept is found by solving the equation f 1x2 = 0, or 0 x 0 = 0. The x-intercept is 0. (c) Use the function to form Table 5 and obtain some points on the graph. Because of the symmetry with respect to the y-axis, we only need to find points 1x, y2 for which x Ú 0. Figure 30 shows the graph of f1x2 = 0 x 0 . Table 5 x y = f (x) = ∣ x ∣ (x, y) 0 0 (0, 0) 1 1 (1, 1) 2 2 (2, 2) 3 3 (3, 3) y 3 (3, 3) (2, 2) (1, 1) 3 2 1 (3, 3) (2, 2) 2 1 1 (1, 1) 1 2 (0, 0) 3 Figure 30 Absolute Value Function x r From the results of Example 2 and Figure 30, we have the following properties of the absolute value function. SECTION 3.4 Library of Functions; Piecewise-defined Functions 239 Properties of f(x) = ∣ x ∣ 1. The domain is the set of all real numbers. The range of f is 5 y y Ú 06 . 2. The x-intercept of the graph of f 1x2 = 0 x 0 is 0. The y-intercept of the graph of f 1x2 = 0 x 0 is also 0. 3. The function is even. The graph is symmetric with respect to the y-axis. 4. The function is decreasing on the interval 1 - q , 02. It is increasing on the interval 10, q 2. 5. The function has an absolute minimum of 0 at x = 0. Seeing the Concept Graph y = 0 x 0 on a square screen and compare what you see with Figure 30. Note that some graphing calculators use abs(x) for absolute value. Below is a list of the key functions that we have discussed. In going through this list, pay special attention to the properties of each function, particularly to the shape of each graph. Knowing these graphs, along with key points on each graph, will lay the foundation for further graphing techniques. y Constant Function f(x) = b (0,b) f1x2 = b b is a real number x Figure 31 Constant Function See Figure 31. The domain of a constant function is the set of all real numbers; its range is the set consisting of a single number b. Its graph is a horizontal line whose y-intercept is b. The constant function is an even function. f (x ) = x y 3 Identity Function f1x2 = x (1, 1) (0, 0) –3 (– 1, –1) 3 x See Figure 32. The domain and the range of the identity function are the set of all real numbers. Its graph is a line whose slope is 1 and whose y-intercept is 0. The line consists of all points for which the x-coordinate equals the y-coordinate. The identity function is an odd function that is increasing over its domain. Note that the graph bisects quadrants I and III. Figure 32 Identity Function y (–2, 4) 4 (–1, 1) –4 f (x ) = x 2 Square Function (2, 4) f1x2 = x2 (1, 1) (0, 0) Figure 33 Square Function 4 x See Figure 33. The domain of the square function is the set of all real numbers; its range is the set of nonnegative real numbers. The graph of this function is a parabola whose intercept is at 10, 02. The square function is an even function that is decreasing on the interval 1 - q , 02 and increasing on the interval 10, q 2. 240 CHAPTER 3 Functions and Their Graphs y Cube Function 4 f(x) = x 3 f1x2 = x3 (1, 1) 4 (1, 1) (0, 0) x 4 See Figure 34. The domain and the range of the cube function are the set of all real numbers. The intercept of the graph is at 10, 02. The cube function is odd and is increasing on the interval 1 - q , q 2. 4 Figure 34 Cube Function y Square Root Function f(x) = x 2 (1, 1) 1 f1x2 = 2x (4, 2) 5 x (0, 0) Figure 35 Square Root Function y 3 3 ( 1–8 , 1–2 (1, 1) ) (2, 2 ) See Figure 35. The domain and the range of the square root function are the set of nonnegative real numbers. The intercept of the graph is at 10, 02. The square root function is neither even nor odd and is increasing on the interval 10, q 2. Cube Root Function 3 f 1x2 = 2 x ( 1–8 , 1–2) 3 3 x (0, 0) 3 (1, 1) (2, 2 ) 3 Figure 36 Cube Root Function y 2 Reciprocal Function (1–2 , 2) f(x ) = (2, 1–2 ) See Figure 36. The domain and the range of the cube root function are the set of all real numbers. The intercept of the graph is at 10, 02. The cube root function is an odd function that is increasing on the interval 1 - q , q 2. f1x2 = 1 –– x 1 x (1, 1) 2 2 x (1, 1) 2 Figure 37 Reciprocal Function 1 Refer to Example 12, page 164, for a discussion of the equation y = . See x Figure 37. The domain and the range of the reciprocal function are the set of all nonzero real numbers. The graph has no intercepts. The reciprocal function is decreasing on the intervals 1 - q , 02 and 10, q 2 and is an odd function. Absolute Value Function y f(x) = ⏐x ⏐ f1x2 = 0 x 0 3 (2, 2) (2, 2) (1, 1) 3 (0, 0) (1, 1) 3 x Figure 38 Absolute Value Function See Figure 38. The domain of the absolute value function is the set of all real numbers; its range is the set of nonnegative real numbers. The intercept of the graph is at 10, 02. If x Ú 0, then f 1x2 = x, and the graph of f is part of the line y = x; if x 6 0, then f1x2 = - x, and the graph of f is part of the line y = - x. The absolute value function is an even function; it is decreasing on the interval 1 - q , 02 and increasing on the interval 10, q 2. SECTION 3.4 Library of Functions; Piecewise-defined Functions 241 The notation int1x2 stands for the largest integer less than or equal to x. For example, 1 3 int112 = 1, int12.52 = 2, inta b = 0, inta - b = - 1, int1p2 = 3 2 4 This type of correspondence occurs frequently enough in mathematics that we give it a name. Table 6 y = f (x) x (x, y) = int (x) DEFINITION Greatest Integer Function -1 -1 ( - 1, - 1) 1 2 -1 1 a - , - 1b 2 1 4 -1 1 a - , - 1b 4 0 0 (0, 0) 1 4 0 1 a , 0b 4 1 2 0 1 a , 0b 2 3 4 0 3 a , 0b 4 - y 4 2 2 2 4 x 3 Figure 39 Greatest Integer Function f1x2 = int1x2* = greatest integer less than or equal to x We obtain the graph of f1x2 = int1x2 by plotting several points. See Table 6. For values of x, - 1 … x 6 0, the value of f 1x2 = int 1x2 is - 1; for values of x, 0 … x 6 1, the value of f is 0. See Figure 39 for the graph. The domain of the greatest integer function is the set of all real numbers; its range is the set of integers. The y-intercept of the graph is 0. The x-intercepts lie in the interval 3 0, 12. The greatest integer function is neither even nor odd. It is constant on every interval of the form 3 k, k + 12, for k an integer. In Figure 39, a solid dot is used to indicate, for example, that at x = 1 the value of f is f112 = 1; an open circle is used to illustrate that the function does not assume the value of 0 at x = 1. Although a precise definition requires the idea of a limit (discussed in calculus), in a rough sense, a function is said to be continuous if its graph has no gaps or holes and can be drawn without lifting a pencil from the paper on which the graph is drawn. We contrast this with a discontinuous function. A function is discontinuous if its graph has gaps or holes and so cannot be drawn without lifting a pencil from the paper. From the graph of the greatest integer function, we can see why it is also called a step function. At x = 0, x = {1, x = {2, and so on, this function is discontinuous because, at integer values, the graph suddenly “steps” from one value to another without taking on any of the intermediate values. For example, to the immediate left of x = 3, the y-coordinates of the points on the graph are 2, and at x = 3 and to the immediate right of x = 3, the y-coordinates of the points on the graph are 3. Consequently, the graph has gaps in it. COMMENT When graphing a function using a graphing utility, typically you can choose either connected mode, in which points plotted on the screen are connected, making the graph appear without any breaks, or dot mode, in which only the points plotted appear. When graphing the greatest integer function with a graphing utility, it may be necessary to be in dot mode. This is to prevent the utility from “connecting the dots” when f1x2 changes from one integer value to the next. However, some utilities will display the gaps even when in “connected” mode. See Figure 40. ■ 6 6 6 22 6 22 Figure 40 f 1x2 = int 1x2 D TI-83 Plus, connected mode 22 6 22 E TI-83 Plus, dot mode *Some texts use the notation f 1x2 = 3 x 4 instead of int 1x2. −2 6 −2 (c) TI-84 Plus C 242 CHAPTER 3 Functions and Their Graphs The functions discussed so far are basic. Whenever you encounter one of them, you should see a mental picture of its graph. For example, if you encounter the function f1x2 = x2, you should see in your mind’s eye a picture like Figure 33. Now Work PROBLEMS 11 THROUGH 18 2 Graph Piecewise-defined Functions Sometimes a function is defined using different equations on different parts of its domain. For example, the absolute value function f1x2 = 0 x 0 is actually defined by two equations: f 1x2 = x if x Ú 0 and f1x2 = - x if x 6 0. For convenience, these equations are generally combined into one expression as f1x2 = 0 x 0 = e x if x Ú 0 - x if x 6 0 When a function is defined by different equations on different parts of its domain, it is called a piecewise-defined function. EX AMPLE 3 Analyzing a Piecewise-defined Function The function f is defined as - 2x + 1 if - 3 … x 6 1 f1x2 = c 2 if x = 1 x2 if x 7 1 (a) Find f1 - 22 , f112, and f122. (c) Locate any intercepts. (e) Use the graph to find the range of f. Solution (b) Determine the domain of f. (d) Graph f . (f) Is f continuous on its domain? (a) To find f1 - 22, observe that when x = - 2, the equation for f is given by f1x2 = - 2x + 1. So f1 - 22 = - 2( - 2) + 1 = 5 When x = 1, the equation for f is f 1x2 = 2. So, f112 = 2 When x = 2, the equation for f is f 1x2 = x2. So f122 = 22 = 4 (b) To find the domain of f, look at its definition. Since f is defined for all x greater than or equal to - 3, the domain of f is 5 x x Ú - 36 , or the interval 3 - 3, q 2 . (c) The y-intercept of the graph of the function is f 102 . Because the equation for f when x = 0 is f1x2 = - 2x + 1, the y-intercept is f102 = - 2102 + 1 = 1. The x-intercepts of the graph of a function f are the real solutions to the equation f1x2 = 0. To find the x-intercepts of f, solve f1x2 = 0 for each “piece” of the function, and then determine what values of x, if any, satisfy the condition that defines the piece. f1x2 = 0 - 2x + 1 = 0 -3 … x 6 1 - 2x = - 1 1 x = 2 f1x2 = 0 2 = 0 x = 1 No solution f1x2 = 0 x2 = 0 x = 0 x 7 1 SECTION 3.4 Library of Functions; Piecewise-defined Functions y 8 4 (1,2) (2,4) (0,1) 4 ( 1–2 , 0) (1, x 1) 243 1 The first potential x-intercept, x = , satisfies the condition - 3 … x 6 1, so 2 1 x = is an x-intercept. The second potential x-intercept, x = 0, does not satisfy 2 1 the condition x 7 1, so x = 0 is not an x-intercept. The only x-intercept is . The 2 1 intercepts are (0, 1) and a , 0b . 2 (d) To graph f , graph each “piece.” First graph the line y = - 2x + 1 and keep only the part for which - 3 … x 6 1. Then plot the point 11, 22 because, when x = 1, f1x2 = 2. Finally, graph the parabola y = x2 and keep only the part for which x 7 1. See Figure 41. (e) From the graph, we conclude that the range of f is 5 y y 7 - 16 , or the interval 1 - 1, q 2. (f) The function f is not continuous because there is a “jump” in the graph at x = 1. r Figure 41 Now Work EXAMPL E 4 PROBLEM 31 Cost of Electricity In the spring of 2014, Duke Energy Progress supplied electricity to residences in South Carolina for a monthly customer charge of $6.50 plus 9.971¢ per kilowatt-hour (kWh) for the first 800 kWh supplied in the month and 8.971¢ per kWh for all usage over 800 kWh in the month. (a) What is the charge for using 300 kWh in a month? (b) What is the charge for using 1500 kWh in a month? (c) If C is the monthly charge for x kWh, develop a model relating the monthly charge and kilowatt-hours used. That is, express C as a function of x. Source: Duke Energy Progress, 2014 Solution (a) For 300 kWh, the charge is $6.50 plus (9.971c = $0.09971) per kWh. That is, Charge = $6.50 + $0.09971(300) = $36.41 (b) For 1500 kWh, the charge is $6.50 plus 9.971c per kWh for the first 800 kWh plus 8.971c per kWh for the 700 in excess of 800. That is, Charge = $6.50 + $0.09971(800) + $0.08971(700) = $ 149.07 (c) Let x represent the number of kilowatt-hours used. If 0 … x … 800, then the monthly charge C (in dollars) can be found by multiplying x times $0.09971and adding the monthly customer charge of $6.50. So if 0 … x … 800, then C(x) = 0.09971x + 6.50 For x 7 800, the charge is 0.09971(800) + 6.50 + 0.08971(x - 800), since (x - 800) equals the usage in excess of 800 kWh, which costs $0.08971 per kWh. That is, if x 7 800, then C C(x) = 0.09971(800) + 6.50 + 0.08971(x - 800) = 79.768 + 6.50 + 0.08971x - 71.768 = 0.08971x + 14.50 Charge (dollars) 180 (1500, 149.07) 120 60 The rule for computing C follows two equations: (800, 86.27) (0, 6.50) C 1x2 = e (300, 36.41) 400 Figure 42 800 1200 Usage (kWh) x 0.09971x + 6.50 if 0 … x … 800 0.08971x + 14.50 if x 7 800 The Model See Figure 42 for the graph. Note that the two “pieces” are linear, but they have different slopes (rates), and meet at the point (800, 86.27). r 244 CHAPTER 3 Functions and Their Graphs 3.4 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Sketch the graph of y = 1x. (p. 163) 2. Sketch the graph of y = 3. List the intercepts of the equation y = x3 - 8. (pp. 159–160) 1 . (p. 164) x Concepts and Vocabulary 4. The function f 1x2 = x2 is decreasing on the interval . 5. When functions are defined by more than one equation, functions. they are called 6. True or False The cube function is odd and is increasing on the interval 1 - q , q 2. 9. Which of the following functions has a graph that is symmetric about the y-axis? 1 (a) y = 2x (b) y = x (c) y = x3 (d) y = x 10. Consider the following function. 3x - 2 f(x) = c x2 + 5 3 7. True or False The cube root function is odd and is decreasing on the interval 1 - q , q 2. 8. True or False The domain and the range of the reciprocal function are the set of all real numbers. x 6 2 2 … x 6 10 x Ú 10 if if if Which “piece(s)” should be used to find the y-intercept? (a) 3x - 2 (b) x2 + 5 (c) 3 (d) all three Skill Building In Problems 11–18, match each graph to its function. A. Constant function E. Square root function B. Identity function F. Reciprocal function C. Square function G. Absolute value function D. Cube function H. Cube root function 11. 12. 13. 14. 15. 16. 17. 18. In Problems 19–26, sketch the graph of each function. Be sure to label three points on the graph. 19. f 1x2 = x 20. f 1x2 = x2 21. f 1x2 = x3 22. f 1x2 = 1x 23. f 1x2 = 24. f 1x2 = 0 x 0 3 25. f 1x2 = 2x 26. f 1x2 = 3 1 x x2 27. If f 1x2 = c 2 2x + 1 find: (a) f 1 - 22 29. If f 1x2 = e 2x - 4 x3 - 2 find: (a) f 102 if x 6 0 if x = 0 if x 7 0 (b) f 102 - 3x 28. If f 1x2 = c 0 2x2 + 1 (c) f 122 if - 1 … x … 2 if 2 6 x … 3 (b) f 112 find: (a) f 1 - 22 30. If f 1x2 = e (c) f 122 (d) f 132 x3 3x + 2 find: (a) f 1 - 12 if x 6 - 1 if x = - 1 if x 7 - 1 (b) f 1 - 12 (c) f 102 if - 2 … x 6 1 if 1 … x … 4 (b) f 102 (c) f 112 (d) f(3) In Problems 31–42: (a) Find the domain of each function. (d) Based on the graph, find the range. 31. f 1x2 = b 2x 1 if x ≠ 0 if x = 0 (b) Locate any intercepts. (e) Is f continuous on its domain? 32. f 1x2 = b 3x 4 if x ≠ 0 if x = 0 (c) Graph each function. 33. f 1x2 = b - 2x + 3 3x - 2 if x 6 1 if x Ú 1 SECTION 3.4 Library of Functions; Piecewise-defined Functions 34. f 1x2 = b x + 3 - 2x - 3 37. f 1x2 = b 1 + x x2 40. f 1x2 = b x + 3 35. f 1x2 = c 5 -x + 2 if x 6 - 2 if x Ú - 2 1 if x 6 0 if x Ú 0 38. f 1x2 = c x 3 2 x 2 - x if - 3 … x 6 1 2x if x 7 1 41. f 1x2 = 2 int 1x2 if - 2 … x 6 1 if x = 1 if x 7 1 if x 6 0 if - 3 … x 6 0 if x = 0 if x 7 0 2x + 5 36. f 1x2 = c - 3 - 5x 39. f 1x2 = b if x Ú 0 0x0 x 245 if - 2 … x 6 0 3 if x 7 0 42. f 1x2 = int 12x2 In Problems 43–46, the graph of a piecewise-defined function is given. Write a definition for each function. y 43. y 2 44. 2 (2, 2) (1, 1) (1, 1) 2 (0, 0) 2 x 2 y (0, 2) 46. 2 (2, 1) (2, 1) (1, 1) y 45. (0, 0) 2 x 2 (1, 0) (0, 0) (2, 0) x 2 (1, 1) 2 x (1, 1) 47. If f 1x2 = int 12x2, find (a) f 11.22 (b) f 11.62 (c) f 1 - 1.82 x 48. If f 1x2 = int a b, find 2 (a) f 11.22 (b) f 11.62 (c) f 1 - 1.82 Applications and Extensions 49. Tablet Service Sprint offers a monthly tablet plan for $34.99. It includes 3 gigabytes of data and charges $15 per gigabyte for additional gigabytes. The following function is used to compute the monthly cost for a subscriber. 34.99 C(x) = e 15x - 10.01 if if 0 … x … 3 x 7 3 Compute the monthly cost for each of the following gigabytes of use. (a) 2 (b) 5 (c) 13 Source: Sprint, April 2014 50. Parking at O’Hare International Airport The short-term (no more than 24 hours) parking fee F (in dollars) for parking x hours on a weekday at O’Hare International Airport’s main parking garage can be modeled by the function 3 F 1x2 = c 5 int 1x + 12 + 1 50 if 0 6 x … 3 if 3 6 x 6 9 if 9 … x … 24 Determine the fee for parking in the short-term parking garage for (a) 2 hours (b) 7 hours (c) 15 hours (d) 8 hours and 24 minutes Source: O’Hare International Airport 51. Cost of Natural Gas In March 2014, Laclede Gas had the rate schedule (on, right) for natural gas usage in single-family residences. (a) What is the charge for using 20 therms in a month? (b) What is the charge for using 150 therms in a month? (c) Develop a function that models the monthly charge C for x therms of gas. (d) Graph the function found in part (c). Monthly service charge Delivery charge First 30 therms Over 30 therms Natural gas cost First 30 therms Over 30 therms $19.50 $0.91686/therm $0 $0.3313/therm $0.5757/therm Source: Laclede Gas 52. Cost of Natural Gas In April 2014, Nicor Gas had the following rate schedule for natural gas usage in small businesses. Monthly customer charge Distribution charge 1st 150 therms Next 4850 therms Over 5000 therms Gas supply charge $72.60 $0.1201/therm $0.0549/therm $0.0482/therm $0.68/therm (a) What is the charge for using 1000 therms in a month? (b) What is the charge for using 6000 therms in a month? (c) Develop a function that models the monthly charge C for x therms of gas. (d) Graph the function found in part (c). Source: Nicor Gas, 2014 246 CHAPTER 3 Functions and Their Graphs 53. Federal Income Tax Two 2014 Tax Rate Schedules are given in the accompanying table. If x equals taxable income and y equals the tax due, construct a function y = f 1x2 for Schedule X. 2014 Tax Rate Schedules Schedule X—Single If Taxable Income is Over Schedule Y-1—Married Filing Jointly or Qualified Widow(er) But Not Over The Tax is This Amount $0 $9,075 $0 + 10% $0 9,075 36,900 907.50 + 15% 9,075 Plus This % Of the Excess Over If Taxable Income is Over But Not Over The Tax is This Amount $0 $18,150 $0 + 10% $0 18,150 73,800 1,815 + 15% 18,150 Plus This % Of the Excess Over 36,900 89,350 5,081.25 + 25% 36,900 73,800 148,850 10,162.50 + 25% 73,800 89,350 186,350 18,193.75 + 28% 89,350 148,850 226,850 28,925.00 + 28% 148,850 186,350 405,100 45,353.75 + 33% 186,350 226,850 405,100 50,765.00 + 33% 226,850 405,100 406,750 117,541.25 + 35% 405,100 405,100 457,600 109,587.50 + 35% 405,100 406,750 - 118,188.75 + 39.6% 406,750 457,600 - 127,962.50 + 39.6% 457,600 54. Federal Income Tax Refer to the 2014 tax rate schedules. If x equals taxable income and y equals the tax due, construct a function y = f 1x2 for Schedule Y-1. 55. Cost of Transporting Goods A trucking company transports goods between Chicago and New York, a distance of 960 miles. The company’s policy is to charge, for each pound, $0.50 per mile for the first 100 miles, $0.40 per mile for the next 300 miles, $0.25 per mile for the next 400 miles, and no charge for the remaining 160 miles. (a) Graph the relationship between the cost of transportation in dollars and mileage over the entire 960-mile route. (b) Find the cost as a function of mileage for hauls between 100 and 400 miles from Chicago. (c) Find the cost as a function of mileage for hauls between 400 and 800 miles from Chicago. 56. Car Rental Costs An economy car rented in Florida from Enterprise® on a weekly basis costs $185 per week. Extra days cost $37 per day until the day rate exceeds the weekly rate, in which case the weekly rate applies. Also, any part of a day used counts as a full day. Find the cost C of renting an economy car as a function of the number x of days used, where 7 … x … 14. Graph this function. 57. Mortgage Fees Fannie Mae charges a loan-level price adjustment (LLPA) on all mortgages, which represents a fee homebuyers seeking a loan must pay. The rate paid depends on the credit score of the borrower, the amount borrowed, and the loan-to-value (LTV) ratio. The LTV ratio is the ratio of amount borrowed to appraised value of the home. For example, a homebuyer who wishes to borrow $250,000 with a credit score of 730 and an LTV ratio of 80% will pay 0.5% (0.005) of $250,000, or $1250. The table shows the LLPA for various credit scores and an LTV ratio of 80%. Credit Score Loan-Level Price Adjustment Rate … 659 3.00% 660–679 2.50% 680–699 1.75% 700–719 1% 720–739 0.5% Ú 740 0.25% Source: Fannie Mae. (a) Construct a function C = C(s), where C is the loan-level price adjustment (LLPA) and s is the credit score of an individual who wishes to borrow $300,000 with an 80% LTV ratio. (b) What is the LLPA on a $300,000 loan with an 80% LTV ratio for a borrower whose credit score is 725? (c) What is the LLPA on a $300,000 loan with an 80% LTV ratio for a borrower whose credit score is 670? 58. Minimum Payments for Credit Cards Holders of credit cards issued by banks, department stores, oil companies, and so on receive bills each month that state minimum amounts that must be paid by a certain due date. The minimum due depends on the total amount owed. One such credit card company uses the following rules: For a bill of less than $10, the entire amount is due. For a bill of at least $10 but less than $500, the minimum due is $10. A minimum of $30 is due on a bill of at least $500 but less than $1000, a minimum of $50 is due on a bill of at least $1000 but less than $1500, and a minimum of $70 is due on bills of $1500 or more. Find the function f that describes the minimum payment due on a bill of x dollars. Graph f. 59. Wind Chill The wind chill factor represents the air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is t W = d 33 - 110.45 + 101v - v2 133 - t2 22.04 33 - 1.5958133 - t2 0 … v 6 1.79 1.79 … v … 20 v 7 20 where v represents the wind speed (in meters per second) and t represents the air temperature (°C). Compute the wind chill for the following: (a) An air temperature of 10°C and a wind speed of 1 meter per second 1m/sec2 (b) An air temperature of 10°C and a wind speed of 5 m/sec (c) An air temperature of 10°C and a wind speed of 15 m/sec (d) An air temperature of 10°C and a wind speed of 25 m/sec (e) Explain the physical meaning of the equation corresponding to 0 … v 6 1.79. (f) Explain the physical meaning of the equation corresponding to v 7 20. SECTION 3.5 Graphing Techniques: Transformations 60. Wind Chill Redo Problem 59(a)–(d) for an air temperature of - 10°C. 61. First-class Mail In 2014 the U.S. Postal Service charged $0.98 postage for first-class mail retail flats (such as an 8.5" by 11" envelope) weighing up to 1 ounce, plus $0.21 for each 247 additional ounce up to 13 ounces. First-class rates do not apply to flats weighing more than 13 ounces. Develop a model that relates C, the first-class postage charged, for a flat weighing x ounces. Graph the function. Source: United States Postal Service Explaining Concepts: Discussion and Writing In Problems 62–69, use a graphing utility. 62. Exploration Graph y = x2. Then on the same screen graph y = x2 + 2, followed by y = x2 + 4, followed by y = x2 - 2. What pattern do you observe? Can you predict the graph of y = x2 - 4? Of y = x2 + 5? 63. Exploration Graph y = x2. Then on the same screen graph y = 1x - 22 2, followed by y = 1x - 42 2, followed by y = 1x + 22 2. What pattern do you observe? Can you predict the graph of y = 1x + 42 2? Of y = 1x - 52 2? 64. Exploration Graph y = 0 x 0 . Then on the same screen graph 1 y = 2 0 x 0 , followed by y = 4 0 x 0 , followed by y = 0 x 0 . 2 What pattern do you observe? Can you predict the graph of 1 y = 0 x 0 ? Of y = 5 0 x 0 ? 4 65. Exploration Graph y = x2. Then on the same screen graph y = - x2. Now try y = 0 x 0 and y = - 0 x 0 . What do you conclude? 66. Exploration Graph y = 1x. Then on the same screen graph y = 1 - x. Now try y = 2x + 1 and y = 21 - x2 + 1. What do you conclude? 67. Exploration Graph y = x3. Then on the same screen graph y = 1x - 12 3 + 2. Could you have predicted the result? 68. Exploration Graph y = x2, y = x4, and y = x6 on the same screen. What do you notice is the same about each graph? What do you notice is different? 69. Exploration Graph y = x3, y = x5, and y = x7 on the same screen. What do you notice is the same about each graph? What do you notice is different? 70. Consider the equation y = b 1 0 if x is rational if x is irrational Is this a function? What is its domain? What is its range? What is its y-intercept, if any? What are its x-intercepts, if any? Is it even, odd, or neither? How would you describe its graph? 71. Define some functions that pass through 10, 02 and 11, 12 and are increasing for x Ú 0. Begin your list with y = 1x, y = x, and y = x2. Can you propose a general result about such functions? Retain Your Knowledge Problems 72–75 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 72. Simplify: 13 + 2i2 14 - 5i2 73. Find the center and radius of the circle x + y = 6y + 16. 2 2 74. Solve: 4x - 512x - 12 = 4 - 71x + 12 75. Ethan has $60,000 to invest. He puts part of the money in a CD that earns 3% simple interest per year and the rest in a mutual fund that earns 8% simple interest per year. How much did he invest in each if his earned interest the first year was $3700? ‘Are You Prepared?’ Answers 1. y 2. 2 (1, 1) y 2 (4, 2) 3. 10, - 82, 12, 02 (1, 1) 2 x (0, 0) 4 x (1, 1) 3.5 Graphing Techniques: Transformations OBJECTIVES 1 Graph Functions Using Vertical and Horizontal Shifts (p. 248) 2 Graph Functions Using Compressions and Stretches (p. 251) 3 Graph Functions Using Reflections about the x-Axis and the y-Axis (p. 253) At this stage, if you were asked to graph any of the functions defined by y = x, 1 3 y = x2, y = x3, y = 2x, y = 2 x, y = , or y = 0 x 0 , your response should be, x “Yes, I recognize these functions and know the general shapes of their graphs.” (If this is not your answer, review the previous section, Figures 32 through 38.) 248 CHAPTER 3 Functions and Their Graphs Sometimes we are asked to graph a function that is “almost” like one that we already know how to graph. In this section, we develop techniques for graphing such functions. Collectively, these techniques are referred to as transformations. 1 Graph Functions Using Vertical and Horizontal Shifts EX AMPLE 1 Vertical Shift Up Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 + 3. Find the domain and range of g. Solution Begin by obtaining some points on the graphs of f and g. For example, when x = 0, then y = f102 = 0 and y = g102 = 3. When x = 1, then y = f 112 = 1 and y = g112 = 4. Table 7 lists these and a few other points on each graph. Notice that each y-coordinate of a point on the graph of g is 3 units larger than the y-coordinate of the corresponding point on the graph of f . We conclude that the graph of g is identical to that of f, except that it is shifted vertically up 3 units. See Figure 43. y = x2 + 3 y (2, 7) (2, 7) Table 7 x y = f (x) = x2 y = g (x) = x2 + 3 -2 4 7 -1 1 4 0 0 3 1 1 4 2 4 7 (1, 4) (1, 4) 5 (2, 4) Up 3 units (2, 4) (0, 3) y = x 2 (1, 1) (1, 1) 3 (0, 0) 3 x Figure 43 The domain of g is all real numbers, or 1 - q , q 2 . The range of g is 3 3, q 2 . EX AMPLE 2 r Vertical Shift Down Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 - 4. Find the domain and range of g. Solution Table 8 lists some points on the graphs of f and g. Notice that each y-coordinate of g is 4 units less than the corresponding y-coordinate of f. To obtain the graph of g from the graph of f, subtract 4 from each y-coordinate on the graph of f. The graph of g is identical to that of f, except that it is shifted down 4 units. See Figure 44. y (– 2, 4) Table 8 x y = f (x) = x2 y = g (x) = x2 − 4 -2 4 0 -1 1 -3 0 0 -4 1 1 -3 2 4 0 y = x2 4 (2, 4) Down 4 units Down 4 units (2, 0) (0, 0) (2, 0) 4 x y = x2 4 5 (0, 4) Figure 44 The domain of g is all real numbers, or 1 - q , q 2 . The range of g is 3 - 4, q 2 . r Note that a vertical shift affects only the range of a function, not the domain. For example, the range of f(x) = x2 is 3 0, q ). In Example 1 the range of g is 3 3, q ), whereas in Example 2 the range of g is 3 - 4, q ). The domain for all three functions is all real numbers. SECTION 3.5 Graphing Techniques: Transformations Exploration Y2 = x 2 + 2 249 On the same screen, graph each of the following functions: Y1 = x2 6 Y2 = x2 + 2 Y3 = x2 - 2 Y1 = x2 −6 6 −2 Figure 45 illustrates the graphs. You should have observed a general pattern. With Y1 = x2 on the screen, the graph of Y2 = x2 + 2 is identical to that of Y1 = x2, except that it is shifted vertically up 2 units. The graph of Y3 = x2 - 2 is identical to that of Y1 = x2, except that it is shifted vertically down 2 units. Y3 = x 2 − 2 Figure 45 We are led to the following conclusions: In Words For y = f (x) + k, k 7 0, add k to each y-coordinate on the graph of y = f (x) to shift the graph up k units. For y = f (x) - k, k 7 0, subtract k from each y-coordinate to shift the graph down k units. If a positive real number k is added to the output of a function y = f 1x2 , the graph of the new function y = f 1x2 + k is the graph of f shifted vertically up k units. If a positive real number k is subtracted from the output of a function y = f1x2, the graph of the new function y = f1x2 - k is the graph of f shifted vertically down k units. Now Work EXAMPL E 3 PROBLEM 39 Horizontal Shift to the Right Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 1x - 2. Find the domain and range of g. Solution Table 9 The function g1x2 = 1x - 2 is basically a square root function. Table 9 lists some points on the graphs of f and g. Note that when f1x2 = 0, then x = 0, and when g1x2 = 0, then x = 2. Also, when f1x2 = 2, then x = 4, and when g1x2 = 2, then x = 6. Notice that the x-coordinates on the graph of g are 2 units larger than the corresponding x-coordinates on the graph of f for any given y-coordinate. We conclude that the graph of g is identical to that of f, except that it is shifted horizontally 2 units to the right. See Figure 46. x y = f (x) = 1x x y = g (x) = 1x − 2 0 0 2 0 1 1 3 1 4 2 6 2 9 3 11 3 y 5 y = "x Right 2 units (4, 2) y = "x − 2 (6, 2) (0, 0) (2, 0) 9 x Right 2 units Figure 46 The domain of g is [2, q ) and the range is [0, q ). EXAMPL E 4 r Horizontal Shift to the Left Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 1x + 4. Find the domain and range of g. Solution The function g1x2 = 1x + 4 is basically a square root function. Its graph is the same as that of f, except that it is shifted horizontally 4 units to the left. See Figure 47 on page 250. 250 CHAPTER 3 Functions and Their Graphs y 5 y = "x + 4 Left 4 units (0, 2) (4, 2) (−4, 0) −5 Figure 47 (0, 0) y = "x 5 x Left 4 units The domain of g is [- 4, q ) and the range is [0, q ). Now Work PROBLEM r 43 Note that a horizontal shift affects only the domain of a function, not the range. For example, the domain of f1x2 = 1x is [0, q ). In Example 3 the domain of g is [2, q ), whereas in Example 4 the domain of g is [- 4, q ). The range for all three functions is [0, q ). Exploration Y2 = (x − 3) 2 On the same screen, graph each of the following functions: Y 1 = x2 Y2 = (x - 3)2 6 Y3 = (x + 2)2 −6 6 Y3 = (x + 2) 2 −2 Y1 = x 2 Figure 48 illustrates the graphs. You should have observed the following pattern. With the graph of Y1 = x2 on the screen, the graph of Y2 = (x - 3)2 is identical to that of Y1 = x2, except that it is shifted horizontally to the right 3 units. The graph of Y3 = (x + 2)2 is identical to that of Y1 = x2, except that it is shifted horizontally to the left 2 units. Figure 48 We are led to the following conclusions: If the argument x of a function f is replaced by x - h, h 7 0, the graph of the new function y = f1x - h2 is the graph of f shifted horizontally right h units. In Words For y = f (x - h), h 7 0, add h to each x-coordinate on the graph of y = f (x) to shift the graph right h units. For y = f (x + h), h 7 0, subtract h from each x-coordinate on the graph of y = f (x) to shift the graph left h units. EX AM PL E 5 If the argument x of a function f is replaced by x + h, h 7 0, the graph of the new function y = f1x + h2 is the graph of f shifted horizontally left h units. Observe the distinction between vertical and horizontal shifts. The graph of f1x2 = x3 + 2 is obtained by shifting the graph of y = x3 up 2 units, because we evaluate the cube function first and then add 2. The graph of g(x) = 1x + 22 3 is obtained by shifting the graph of y = x3 left 2 units, because we add 2 to x before we evaluate the cube function. Vertical and horizontal shifts are sometimes combined. Combining Vertical and Horizontal Shifts Graph the function f1x2 = x + 3 - 5. Find the domain and range of f. Solution We graph f in steps. First, note that the rule for f is basically an absolute value function, so begin with the graph of y = x as shown in Figure 49(a). Next, to get the graph of y = x + 3 , shift the graph of y = x horizontally 3 units to the left. See Figure 49(b). Finally, to get the graph of y = x + 3 - 5, shift the graph of y = x + 3 vertically down 5 units. See Figure 49(c). Note the points plotted on each graph. Using key points can be helpful in keeping track of the transformation that has taken place. 251 SECTION 3.5 Graphing Techniques: Transformations y y y 5 5 5 (−1, 2) (−2, 2) (2, 2) (0, 0) (−5, 2) 5 x (−3, 0) x 2 (−1, −3) (−5, −3) Replace x by x + 3; Horizontal shift left 3 units y = 0x 0 Subtract 5: Vertical shift down 5 units y = 0 x + 30 (−3, −5) y = 0 x + 30 − 5 (b) (a) x 2 (c) Figure 49 The domain of f is all real numbers, or 1 - q , q 2. The range of f is [- 5, q ). Check: Graph Y1 = f 1x2 = x + 3 - 5 and compare the graph to Figure 49(c). r In Example 5, if the vertical shift had been done first, followed by the horizontal shift, the final graph would have been the same. Try it for yourself. Now Work PROBLEMS 45 AND 69 2 Graph Functions Using Compressions and Stretches EXAMPL E 6 Vertical Stretch Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 21x. Solution Table 10 To see the relationship between the graphs of f and g, we form Table 10, listing points on each graph. For each x, the y-coordinate of a point on the graph of g is 2 times as large as the corresponding y-coordinate on the graph of f. The graph of f1x2 = 1x is vertically stretched by a factor of 2 to obtain the graph of g1x2 = 21x. For example, 11, 12 is on the graph of f, but 11, 22 is on the graph of g. See Figure 50. y = 2"x x y = f (x) = 1x y = g (x) = 21x y 0 0 0 5 1 1 2 (4, 4) 4 2 4 (4, 2) 9 3 6 (9, 6) (1, 2) (9, 3) (0, 0) 5 (1, 1) Vertical Compression Use the graph of f1x2 = 0 x 0 to obtain the graph of g1x2 = Solution 10 x r Figure 50 EXAMPL E 7 y = "x 1 0x0. 2 1 as large as the 2 corresponding y-coordinate on the graph of f. The graph of f1x2 = 0 x 0 is vertically 1 1 compressed by a factor of to obtain the graph of g1x2 = 0 x 0 . For example, 12, 22 is 2 2 on the graph of f, but 12, 12 is on the graph of g. See Table 11 and Figure 51 on page 252. For each x, the y-coordinate of a point on the graph of g is 252 CHAPTER 3 Functions and Their Graphs Table 11 x In Words y y = g (x) 1 = x 2 y = f (x) = x -2 2 1 -1 1 1 2 0 0 0 1 1 1 2 2 2 1 y =⏐x⏐ 4 y= (2, 2) (2, 2) (– 2, 1) 1– x ⏐ ⏐ 2 (2, 1) 4 (0, 0) 4 x Figure 51 r When the right side of a function y = f1x2 is multiplied by a positive number a, the graph of the new function y = af1x2 is obtained by multiplying each y-coordinate on the graph of y = f1x2 by a. The new graph is a vertically compressed (if 0 6 a 6 1) or a vertically stretched (if a 7 1) version of the graph of y = f1x2. For y = af (x), a 7 0, the factor a is “outside” the function, so it affects the y-coordinates. Multiply each y-coordinate on the graph of y = f (x) by a. Now Work PROBLEM 47 What happens if the argument x of a function y = f 1x2 is multiplied by a positive number a, creating a new function y = f 1ax2? To find the answer, look at the following Exploration. Exploration On the same screen, graph each of the following functions: Y1 = f (x) = 1x Y2 = f (2x) = 12x 1 x 1 Y3 = f a x b = x = A2 2 A2 Create a table of values to explore the relation between the x- and y-coordinates of each function. Result You should have obtained the graphs in Figure 52. Look at Table 12(a). Note that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also, (0.5, 1), (2, 2), and (4.5, 3) are points on the graph of 1 Y2 = 22x. For a given y-coordinate, the x-coordinate on the graph of Y2 is of the x-coordinate on Y1. 2 3 Y2 = √ 2x Table 12 Y1 = √x x Y3 = Ä 2 0 0 Figure 52 4 (a) (b) We conclude that the graph of Y2 = 22x is obtained by multiplying the x-coordinate of each point on 1 the graph of Y1 = 1x by . The graph of Y2 = 22x is the graph of Y1 = 1x compressed horizontally. 2 Look at Table 12(b). Notice that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also x notice that (2, 1), (8, 2), and (18, 3) are points on the graph of Y3 = . For a given y-coordinate, A2 the x-coordinate on the graph of Y3 is 2 times the x-coordinate on Y1. We conclude that the graph of x Y3 = is obtained by multiplying the x-coordinate of each point on the graph of Y1 = 1x by 2. A2 x The graph of Y3 = is the graph of Y1 = 1x stretched horizontally. A2 SECTION 3.5 Graphing Techniques: Transformations 253 Based on the Exploration, we have the following result: In Words If the argument x of a function y = f1x2 is multiplied by a positive number a, then the graph of the new function y = f1ax2 is obtained by multiplying each 1 x-coordinate of y = f1x2 by . A horizontal compression results if a 7 1, and a a horizontal stretch results if 0 6 a 6 1. For y = f (ax), a 7 0, the factor a is “inside” the function, so it affects the x-coordinates. Multiply each x-coordinate on the 1 graph of y = f (x) by . a Let’s look at an example. Graphing Using Stretches and Compressions EXAMPL E 8 The graph of y = f1x2 is given in Figure 53. Use this graph to find the graphs of (a) y = 2f 1x2 (b) y = f 13x2 (a) The graph of y = 2f1x2 is obtained by multiplying each y-coordinate of y = f1x2 by 2. See Figure 54. Solution (b) The graph of y = f13x2 is obtained from the graph of y = f 1x2 by multiplying 1 each x-coordinate of y = f 1x2 by . See Figure 55. 3 y 3 y 1 ( 2 , 1( 1 3 2 5 3 2 2 3 , 1 2 ( 2 3 2 2 52 3 x 1 3 ( ( 3 , 2 2 2 PROBLEMS 63(e) 2 x 2 3 Figure 55 y = f 13x2 AND 3 ( 2 , 1 ( Figure 54 y = 2f 1x2 Now Work ( 6 , 1( ( 56 , 1( 1 1 Figure 53 y = f 1x2 2 1 x ( y ( 52 , 2( 2 ( 52 , 1( y f(x) 2 ( 2 , 2( r (g) 3 Graph Functions Using Reflections about the x-Axis and the y-Axis Reflection about the x-Axis EXAMPL E 9 Graph the function f1x2 = - x2. Find the domain and range of f. y (2, 4) 4 (1, 1) 4 point (x, y) on the graph of y = x2, the point 1x, - y2 is on the graph of y = - x2, as indicated in Table 13. Draw the graph of y = - x2 by reflecting the graph of y = x2 about the x-axis. See Figure 56. (2, 4) Table 13 (1, 1) (1, – 1) (–1, –1) (–2, –4) Solution Begin with the graph of y = x2, as shown in black in Figure 56. For each y = x2 –4 (2, – 4) 4 x x y = x2 y = −x 2 -2 4 -4 -1 1 -1 0 0 0 1 1 -1 2 4 -4 y = –x2 Figure 56 The domain of f is all real numbers, or ( - q , q ). The range of f is ( - q , 0]. r 254 CHAPTER 3 Functions and Their Graphs When the right side of the function y = f1x2 is multiplied by - 1, the graph of the new function y = - f 1x2 is the reflection about the x-axis of the graph of the function y = f 1x2. Now Work EX AM P L E 1 0 PROBLEM 49 Reflection about the y-Axis Graph the function f1x2 = 1- x. Find the domain and range of f. Solution To get the graph of f 1x2 = 1- x, begin with the graph of y = 1x, as shown in Figure 57. For each point 1x, y2 on the graph of y = 1x, the point 1 - x, y2 is on the graph of y = 1- x. Obtain the graph of y = 1- x by reflecting the graph of y = 1x about the y-axis. See Figure 57. y 4 y= y= –x ( – 1, 1) –5 x (4, 2) ( – 4, 2) (1, 1) (0, 0) 5 x Figure 57 In Words For y = - f (x), multiply each y-coordinate on the graph of y = f (x) by - 1. For y = f (- x), multiply each x-coordinate by - 1. The domain of f is ( - q , 0]. The range of f is the set of all nonnegative real numbers, or [0, q ). r When the graph of the function y = f1x2 is known, the graph of the new function y = f1 - x2 is the reflection about the y-axis of the graph of the function y = f1x2. SUMMARY OF GRAPHING TECHNIQUES To Graph: Draw the Graph of f and: Functional Change to f(x) Vertical shifts y = f 1x2 + k, k 7 0 y = f1x2 - k, k 7 0 Raise the graph of f by k units. Lower the graph of f by k units. Add k to f1x2. Subtract k from f1x2. Horizontal shifts y = f1x + h2, h 7 0 y = f1x - h2 , h 7 0 Shift the graph of f to the left h units. Shift the graph of f to the right h units. Replace x by x + h. Replace x by x - h. Multiply each y-coordinate of y = f1x2 by a. Stretch the graph of f vertically if a 7 1. Compress the graph of f vertically if 0 6 a 6 1. 1 Multiply each x-coordinate of y = f1x2 by . a Stretch the graph of f horizontally if 0 6 a 6 1. Compress the graph of f horizontally if a 7 1. Multiply f1x2 by a. Reflection about the x-axis y = - f1x2 Reflect the graph of f about the x-axis. Multiply f1x2 by - 1. Reflection about the y-axis y = f 1 - x2 Reflect the graph of f about the y-axis. Replace x by - x. Compressing or stretching y = af 1x2, a 7 0 y = f 1ax2, a 7 0 Replace x by ax. SECTION 3.5 Graphing Techniques: Transformations 255 Determining the Function Obtained from a Series of Transformations EX AM PL E 11 Find the function that is finally graphed after the following three transformations are applied to the graph of y = 0 x 0 . 1. Shift left 2 units 2. Shift up 3 units 3. Reflect about the y-axis Solution y = 0x + 20 1. Shift left 2 units: Replace x by x + 2. y = 0x + 20 + 3 2. Shift up 3 units: Add 3. y = 0 -x + 20 + 3 3. Reflect about the y-axis: Replace x by - x. Now Work PROBLEM r 27 Combining Graphing Procedures EX AM PL E 1 2 3 + 1. Find the domain and range of f . x - 2 1 It is helpful to write f as f (x) = 3 a b + 1. Now use the following steps to x - 2 obtain the graph of f . Graph the function f1x2 = Solution 1 x STEP 1: y = Reciprocal function 1 3 STEP 2: y = 3 # a b = x x 3 x - 2 3 STEP 4: y = + 1 x - 2 STEP 3: y = Multiply by 3; vertical stretch by a factor of 3. Replace x by x - 2; horizontal shift to the right 2 units. Add 1; vertical shift up 1 unit. See Figure 58. y 4 y 4 (1, 1) (1, 3) 4 x y 4 (3, 3) 3 2, – 2 ( ) 1 2, – 2 ( ) 24 y 4 24 3 4, – 2 ( ) 4 x 4 (21, 21) x (4, 5–2 ) 4 x 24 (1, 22) (21, 23) 24 1 x (a) y 5 –– Figure 58 (3, 4) Multiply by 3; Vertical stretch (1, 23) 24 24 24 3 x (b) y 5 –– Replace x by x 2 2; Horizontal shift right 2 units 3 x –2 (c) y 5 ––– Add 1; Vertical shift up 1 unit 3 x –2 (d) y 5 ––– 1 1 1 The domain of y = is 5 x x ≠ 06 and its range is 5 y y ≠ 06 . Because we x shifted right 2 units and up 1 unit to obtain f , the domain of f is 5 x x ≠ 26 and its range is 5 y y ≠ 16 . r 256 CHAPTER 3 Functions and Their Graphs HINT: Although the order in which transformations are performed can be altered, consider using the following order for consistency: 1. Reflections 2. Compressions and stretches 3. Shifts Other orderings of the steps shown in Example 12 would also result in the graph of f. For example, try this one: STEP 1: y = 1 x Reciprocal function 1 Replace x by x - 2; horizontal shift to the right 2 units. x - 2 3 STEP 3: y = Multiply by 3; vertical stretch by a factor of 3. x - 2 3 STEP 4: y = + 1 Add 1; vertical shift up 1 unit. x - 2 STEP 2: y = Combining Graphing Procedures EX A MPL E 13 Graph the function f1x2 = 21 - x + 2. Find the domain and range of f . Solution Because horizontal shifts require the form x - h, begin by rewriting f1x2 as f1x2 = 21 - x + 2 = 2 - (x - 1) + 2. Now use the following steps. STEP 1: y = 1 x Square root function STEP 2: y = 2 - x Replace x by - x; reflect about the y-axis. STEP 3: y = 2 - (x - 1) = 21 - x Replace x by x - 1; horizontal shift to the right 1 unit. STEP 4: y = 21 - x + 2 Add 2; vertical shift up 2 units. See Figure 59. (1, 1) 25 y 5 y 5 y 5 (4, 2) (a) y 5 (0, 3) (23, 2) (1, 2) (0, 1) 5 x 25 (0, 0) (24, 2) y (23, 4) 5 x Replace x by 2x; Reflect about y-axis (21, 1) (0, 0) (b) y 5 2x 5 x 25 (1, 0) Replace x by x 2 1; (c) y 5 Horizontal shift 5 right 1 unit 5 5 x 25 2(x 2 1) Add 2; (d) y 5 2 x 1 1 Vertical shift up 2 units 12x Figure 59 The domain of f is ( - q , 1] and the range is [2, q ). Now Work PROBLEM 5 x 12x12 r 55 3.5 Assess Your Understanding Concepts and Vocabulary 1. Suppose that the graph of a function f is known. Then the graph of y = f1x - 22 may be obtained by a(n) shift of the graph of f to the a distance of 2 units. 2. Suppose that the graph of a function f is known. Then the graph of y = f 1 - x2 may be obtained by a reflection about the -axis of the graph of the function y = f 1x2. 1 3. True or False The graph of y = g(x) is the graph of 3 y = g(x) vertically stretched by a factor of 3. 4. True or False The graph of y = - f 1x2 is the reflection about the x-axis of the graph of y = f 1x2. 5. Which of the following functions has a graph that is the graph of y = 2x shifted down 3 units? (a) y = 2x + 3 (b) y = 2x - 3 (c) y = 2x + 3 (d) y = 2x - 3 6. Which of the following functions has a graph that is the graph of y = f (x) compressed horizontally by a factor of 4? 1 (a) y = f (4x) (b) y = f ¢ x≤ 4 (c) y = 4f (x) (d) y = 1 f (x) 4 257 SECTION 3.5 Graphing Techniques: Transformations Skill Building In Problems 7–18, match each graph to one of the following functions: A. y = x2 + 2 E. y = 1x - 22 I. y = 2x 7. C. y = 0 x 0 + 2 B. y = - x2 + 2 2 F. y = - 1x + 22 J. y = - 2x 2 y 3 H. y = - 0 x + 2 0 K. y = 2 0 x 0 2 8. D. y = - 0 x 0 + 2 G. y = 0 x - 2 0 2 L. y = - 2 0 x 0 9. y 3 10. y 1 3 3 x 3 11. 12. y 3 3 3x 3 3 x 3 13. y 5 y 3 3 x 14. y 3 3 3 x y 8 3 x 6 3 3 15. 16. y 4 4 4 x 1 3 x 3 3 x 18. y 4 4 3 4 4 3 17. y 3 6 x 4 x y 3 3 4 3 x 3 In Problems 19–26, write the function whose graph is the graph of y = x3, but is: 19. Shifted to the right 4 units 20. Shifted to the left 4 units 21. Shifted up 4 units 22. Shifted down 4 units 23. Reflected about the y-axis 24. Reflected about the x-axis 25. Vertically stretched by a factor of 4 26. Horizontally stretched by a factor of 4 In Problems 27–30, find the function that is finally graphed after each of the following transformations is applied to the graph of y = 1x in the order stated. 27. (1) Shift up 2 units (2) Reflect about the x-axis (3) Reflect about the y-axis 28. (1) Reflect about the x-axis (2) Shift right 3 units (3) Shift down 2 units 29. (1) Reflect about the x-axis (2) Shift up 2 units (3) Shift left 3 units 30. (1) Shift up 2 units (2) Reflect about the y-axis (3) Shift left 3 units 31. If 13, 62 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = - f 1x2? (a) 16, 32 (b) 16, - 32 32. If 13, 62 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = f 1 - x2? (a) 16, 32 (b) 16, - 32 33. If 11, 32 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = 2f 1x2? 3 (a) a1, b (b) 12, 32 2 1 (c) 11, 62 (d) a , 3b 2 34. If 14, 22 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = f 12x2 ? (a) 14, 12 (b) 18, 22 (c) 13, - 62 (d) 1 - 3, 62 (c) 13, - 62 (d) 1 - 3, 62 (c) 12, 22 (d) 14, 42 258 CHAPTER 3 Functions and Their Graphs 35. Suppose that the x-intercepts of the graph of y = f 1x2 are - 5 and 3. (a) What are the x-intercepts of the graph of y = f 1x + 22? (b) What are the x-intercepts of the graph of y = f 1x - 22? (c) What are the x-intercepts of the graph of y = 4f 1x2? (d) What are the x-intercepts of the graph of y = f 1 - x2? 36. Suppose that the x-intercepts of the graph of y = f 1x2 are - 8 and 1. (a) What are the x-intercepts of the graph of y = f 1x + 42? (b) What are the x-intercepts of the graph of y = f 1x - 32? (c) What are the x-intercepts of the graph of y = 2f 1x2? (d) What are the x-intercepts of the graph of y = f 1 - x2? 37. Suppose that the function y = f 1x2 is increasing on the interval 1 - 1, 52. (a) Over what interval is the graph of y = f 1x + 22 increasing? (b) Over what interval is the graph of y = f 1x - 52 increasing? (c) What can be said about the graph of y = - f 1x2? (d) What can be said about the graph of y = f 1 - x2? 38. Suppose that the function y = f 1x2 is decreasing on the interval 1 - 2, 72. (a) Over what interval is the graph of y = f 1x + 22 decreasing? (b) Over what interval is the graph of y = f 1x - 52 decreasing? (c) What can be said about the graph of y = - f 1x2? (d) What can be said about the graph of y = f 1 - x2? In Problems 39–62, graph each function using the techniques of shifting, compressing, stretching, and/or reflecting. Start with the graph of the basic function (for example, y = x2) and show all stages. Be sure to show at least three key points. Find the domain and the range of each function. 39. f 1x2 = x2 - 1 40. f 1x2 = x2 + 4 41. g1x2 = x3 + 1 42. g1x2 = x3 - 1 43. h1x2 = 2x + 2 44. h1x2 = 2x + 1 45. f 1x2 = 1x - 12 3 + 2 46. f 1x2 = 1x + 22 3 - 3 47. g1x2 = 41x 3 49. f 1x2 = - 2 x 50. f 1x2 = - 1x 51. f 1x2 = 21x + 12 2 - 3 52. f 1x2 = 31x - 22 2 + 1 53. g1x2 = 22x - 2 + 1 54. g1x2 = 3 0 x + 1 0 - 3 55. h1x2 = 1 - x - 2 56. h1x2 = 57. f 1x2 = - 1x + 12 3 - 1 58. f 1x2 = - 42x - 1 59. g1x2 = 2 0 1 - x 0 60. g1x2 = 422 - x 61. h1x2 = 48. g1x2 = 1 1x 2 1 2x 4 + 2 x 3 62. h1x2 = 2x - 1 + 3 In Problems 63–66, the graph of a function f is illustrated. Use the graph of f as the first step toward graphing each of the following functions: (a) F 1x2 = f 1x2 + 3 (b) G1x2 = f 1x + 22 (c) P 1x2 = - f 1x2 1 (e) Q1x2 = f 1x2 2 (f) g1x2 = f 1 - x2 (g) h1x2 = f 12x2 y 4 (0, 2) 63. y 4 64. (2, 2) 2 1 (2, 2) (4, 0) 4 (4, 2) 2 x 2 y 65. 4 2 (4, 2) 2 (2, 2) 4 x 2 (4, 2) π π –2 1 π ( –2 , 1) (d) H1x2 = f 1x + 12 - 2 π– 2 y 66. (π–2 , 1) 1 π x π π –2 (π, 1) π π– 1 2 x (π, 1) Mixed Practice In Problems 67–74, complete the square of each quadratic expression. Then graph each function using the technique of shifting. (If necessary, refer to Chapter R, Section R.5 to review completing the square.) 67. f 1x2 = x2 + 2x 71. f 1x2 = 2x2 - 12x + 19 68. f 1x2 = x2 - 6x 72. f 1x2 = 3x2 + 6x + 1 69. f 1x2 = x2 - 8x + 1 73. f 1x2 = - 3x2 - 12x - 17 70. f 1x2 = x2 + 4x + 2 74. f 1x2 = - 2x2 - 12x - 13 SECTION 3.5 Graphing Techniques: Transformations 259 Applications and Extensions 75. The graph of a function f is illustrated in the figure. (a) Draw the graph of y = 0 f 1x2 0 . y 2 (1, 1) 3 (2, 1) (2, 0) 3 x (1, 1) 2 (b) Draw the graph of y = f 1 0 x 0 2. 72 (2, 0) 23 (21, 72) 68 64 (6, 65) 60 56 t 4 8 12 16 20 24 Time (hours after midnight) 84. Digital Music Revenues The total projected worldwide digital music revenues R, in millions of dollars, for the years 2012 through 2017 can be estimated by the function R 1x2 = 28.6x2 + 300x + 4843 (1, 1) (22, 0) 3 x (21, 21) 22 76 0 76. The graph of a function f is illustrated in the figure. (a) Draw the graph of y = 0 f 1x2 0 . y 2 80 Temperature (°) ) (b) Draw the graph of y = f 1 0 x 0 2. T (0, 21) 77. Suppose 11, 32 is a point on the graph of y = f 1x2 . (a) What point is on the graph of y = f 1x + 32 - 5? (b) What point is on the graph of y = - 2f 1x - 22 + 1? (c) What point is on the graph of y = f 12x + 32? 78. Suppose 1 - 3, 52 is a point on the graph of y = g1x2 . (a) What point is on the graph of y = g1x + 12 - 3? (b) What point is on the graph of y = - 3g1x - 42 + 3? (c) What point is on the graph of y = g13x + 92? 79. Graph the following functions using transformations. (b) g(x) = - int(x) (a) f(x) = int( - x) 80. Graph the following functions using transformations (a) f(x) = int(x - 1) (b) g(x) = int(1 - x) 81. (a) Graphf(x) = x - 3 - 3 using transformations. (b) Find the area of the region that is bounded by f and the x-axis and lies below the x-axis. 82. (a) Graph f(x) = - 2 x - 4 + 4 using transformations. (b) Find the area of the region that is bounded by f and the x-axis and lies above the x-axis. 83. Thermostat Control Energy conservation experts estimate that homeowners can save 5% to 10% on winter heating bills by programming their thermostats 5 to 10 degrees lower while sleeping. In the graph (top, right), the temperature T (in degrees Fahrenheit) of a home is given as a function of time t (in hours after midnight) over a 24-hour period. (a) At what temperature is the thermostat set during daytime hours? At what temperature is the thermostat set overnight? (b) The homeowner reprograms the thermostat to y = T 1t2 - 2. Explain how this affects the temperature in the house. Graph this new function. (c) The homeowner reprograms the thermostat to y = T 1t + 12. Explain how this affects the temperature in the house. Graph this new function. Source: Roger Albright, 547 Ways to Be Fuel Smart, 2000 where x is the number of years after 2012. (a) Find R 102, R 132, and R 152 and explain what each value represents. (b) Find r 1x2 = R 1x - 22. (c) Find r 122, r 152, and r 172 and explain what each value represents. (d) In the model r = r 1x2 , what does x represent? (e) Would there be an advantage in using the model r when estimating the projected revenues for a given year instead of the model R ? Source: IFPI Digital Music Report 2013 85. Temperature Measurements The relationship between the Celsius (°C) and Fahrenheit (°F) scales for measuring temperature is given by the equation F = 9 C + 32 5 The relationship between the Celsius (°C) and Kelvin (K) 9 scales is K = C + 273. Graph the equation F = C + 32 5 using degrees Fahrenheit on the y-axis and degrees Celsius on the x-axis. Use the techniques introduced in this section to obtain the graph showing the relationship between Kelvin and Fahrenheit temperatures. 86. Period of a Pendulum The period T (in seconds) of a simple pendulum is a function of its length l (in feet) defined by the equation T = 2p l Ag where g ≈ 32.2 feet per second per second is the acceleration due to gravity. (Continued on page 260.) 260 CHAPTER 3 Functions and Their Graphs (a) Use a graphing utility to graph the function T = T 1l2. (b) Now graph the functions T = T 1l + 12, T = T 1l + 22, and T = T 1l + 32. (c) Discuss how adding to the length l changes the period T. (d) Now graph the functions T = T 12l2, T = T 13l2, and T = T 14l2. (e) Discuss how multiplying the length l by factors of 2, 3, and 4 changes the period T. 87. The equation y = 1x - c2 2 defines a family of parabolas, one parabola for each value of c. On one set of coordinate axes, graph the members of the family for c = 0, c = 3, and c = - 2. 88. Repeat Problem 87 for the family of parabolas y = x2 + c. Explaining Concepts: Discussion and Writing 89. Suppose that the graph of a function f is known. Explain how the graph of y = 4f 1x2 differs from the graph of y = f 14x2. ideas presented in this section, what do you think is the area under the curve of y = 1- x bounded from below by the x-axis and on the left by x = - 4? Justify your answer. 90. Suppose that the graph of a function f is known. Explain how the graph of y = f 1x2 - 2 differs from the graph of y = f 1x - 22. 92. Explain how the range of the function f (x) = x2 compares to the range of g(x) = f (x) + k. 91. The area under the curve y = 1x bounded from below by the 16 square units. Using the x-axis and on the right by x = 4 is 3 93. Explain how the domain of g(x) = 2x compares to the domain of g(x - k), where k Ú 0. Retain Your Knowledge Problems 94–97 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 94. Determine the slope and y-intercept of the graph of 3x - 5y = 30. 1x -2y3 2 4 1x2y -5 2 -2 96. The amount of water used when taking a shower varies directly with the number of minutes the shower is run. If a 4-minute shower uses 7 gallons of water, how much water is used in a 9-minute shower? 95. Simplify 97. List the intercepts and test for symmetry: y2 = x + 4 3.6 Mathematical Models: Building Functions OBJECTIVE 1 Build and Analyze Functions (p. 260) 1 Build and Analyze Functions Real-world problems often result in mathematical models that involve functions. These functions need to be constructed or built based on the information given. In building functions, we must be able to translate the verbal description into the language of mathematics. This is done by assigning symbols to represent the independent and dependent variables and then by finding the function or rule that relates these variables. EX AMPLE 1 Finding the Distance from the Origin to a Point on a Graph Let P = 1x, y2 be a point on the graph of y = x2 - 1. (a) Express the distance d from P to the origin O as a function of x. (b) What is d if x = 0? (c) What is d if x = 1? 22 ? (d) What is d if x = 2 SECTION 3.6 Mathematical Models: Building Functions 261 (e) Use a graphing utility to graph the function d = d 1x2, x Ú 0. Rounding to two decimal places, find the value(s) of x at which d has a local minimum. [This gives the point(s) on the graph of y = x2 - 1 closest to the origin.] Solution y (a) Figure 60 illustrates the graph of y = x2 - 1. The distance d from P to O is 2 d = 2 1x - 02 2 + 1y - 02 2 = 2x2 + y2 1 P5 (x, y) (0, 0) d 1 21 2 x Since P is a point on the graph of y = x2 - 1, substitute x2 - 1 for y. Then d 1x2 = 2x2 + 1x2 - 12 2 = 2x4 - x2 + 1 21 Figure 60 y = x2 - 1 The distance d is expressed as a function of x. (b) If x = 0, the distance d is d 102 = 204 - 02 + 1 = 21 = 1 (c) If x = 1, the distance d is d 112 = 214 - 12 + 1 = 1 (d) If x = 22 , the distance d is 2 da 2 12 12 4 12 2 1 1 13 b = a b - a b + 1 = - + 1 = 2 B 2 2 B4 2 2 (e) Figure 61 shows the graph of Y1 = 2x4 - x2 + 1. Using the MINIMUM 0 2 Figure 61 d(x) = 2x4 - x2 + 1 feature on a graphing utility, we find that when x ≈ 0.71 the value of d is smallest. The local minimum is d ≈ 0.87 rounded to two decimal places. Since d 1x2 is even, it follows by symmetry that when x ≈ - 0.71, the value of d is also a local minimum. Since 1 {0.712 2 - 1 ≈ - 0.50, the points 1 - 0.71, - 0.502 and 10.71, - 0.502 on the graph of y = x2 - 1 are closest to the origin. r Now Work EXAMPL E 2 y 30 20 1 y 25 x 2 1 2 3 4 5 x (0,0) Figure 62 1 Area of a Rectangle A rectangle has one corner in quadrant I on the graph of y = 25 - x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See Figure 62. (x, y) 10 PROBLEM Solution (a) Express the area A of the rectangle as a function of x. (b) What is the domain of A? (c) Graph A = A 1x2. (d) For what value of x is the area largest? (a) The area A of the rectangle is A = xy, where y = 25 - x2. Substituting this expression for y, we obtain A 1x2 = x125 - x2 2 = 25x - x3. (b) Since 1x, y2 is in quadrant I, we have x 7 0. Also, y = 25 - x2 7 0, which implies that x2 6 25, so - 5 6 x 6 5. Combining these restrictions, we have the domain of A as 5 x 0 6 x 6 56 , or 10, 52 using interval notation. (c) See Figure 63 on page 262 for the graph of A = A 1x2. (d) Using MAXIMUM, we find that the maximum area is 48.11 square units at x = 2.89 units, each rounded to two decimal places. See Figure 64. 262 CHAPTER 3 Functions and Their Graphs 50 50 5 0 0 Figure 63 A(x) = 25x - x3 Now Work PROBLEM 5 0 0 Figure 64 r 7 Close Call? EX AMPLE 3 Suppose two planes flying at the same altitude are headed toward each other. One plane is flying due south at a groundspeed of 400 miles per hour and is 600 miles from the potential intersection point of the planes. The other plane is flying due west with a groundspeed of 250 miles per hour and is 400 miles from the potential intersection point of the planes. See Figure 65. (a) Build a model that expresses the distance d between the planes as a function of time t. (b) Use a graphing utility to graph d = d(t). How close do the planes come to each other? At what time are the planes closest? Solution N Plane 600 miles 400 mph (a) Refer to Figure 65. The distance d between the two planes is the hypotenuse of a right triangle. At any time t, the length of the north/south leg of the triangle is 600 - 400t. At any time t, the length of the east/west leg of the triangle is 400 - 250t. Use the Pythagorean Theorem to find that the square of the distance between the two planes is d d 2 = (600 - 400t)2 + (400 - 250t)2 Plane 250 mph E 400 miles Therefore, the distance between the two planes as a function of time is given by the model d(t) = 2(600 - 400t)2 + (400 - 250t)2 Figure 65 (b) Figure 66(a) shows the graph of d = d(t). Using MINIMUM, the minimum distance between the planes is 21.20 miles, and the time at which the planes are closest is after 1.53 hours, each rounded to two decimal places. See Figure 66(b). 500 2 0 −50 Figure 66 (b) (a) Now Work PROBLEM 19 r SECTION 3.6 Mathematical Models: Building Functions 263 3.6 Assess Your Understanding Applications and Extensions 1. Let P = 1x, y2 be a point on the graph of y = x2 - 8. (a) Express the distance d from P to the origin as a function of x. (b) What is d if x = 0? (c) What is d if x = 1? (d) Use a graphing utility to graph d = d1x2. (e) For what values of x is d smallest? 2. Let P = 1x, y2 be a point on the graph of y = x2 - 8. (a) Express the distance d from P to the point 10, - 12 as a function of x. (b) What is d if x = 0? (c) What is d if x = - 1? (d) Use a graphing utility to graph d = d1x2. (e) For what values of x is d smallest? 3. Let P = 1x, y2 be a point on the graph of y = 1x. (a) Express the distance d from P to the point 11, 02 as a function of x. (b) Use a graphing utility to graph d = d1x2. (c) For what values of x is d smallest? 4. Let P = 1x, y2 be a point on the graph of y = y y 16 x 2 16 (x, y) 8 x 4 (0,0) 8. A rectangle is inscribed in a semicircle of radius 2. See the figure. Let P = 1x, y2 be the point in quadrant I that is a vertex of the rectangle and is on the circle. y y 4 x2 P (x, y ) 2 2 x 1 . x (a) Express the distance d from P to the origin as a function of x. (b) Use a graphing utility to graph d = d1x2. (c) For what values of x is d smallest? (a) Express the area A of the rectangle as a function of x. (b) Express the perimeter p of the rectangle as a function of x. (c) Graph A = A1x2. For what value of x is A largest? (d) Graph p = p1x2. For what value of x is p largest? 5. A right triangle has one vertex on the graph of y = x3, x 7 0, at 1x, y2, another at the origin, and the third on the positive y-axis at 10, y2, as shown in the figure. Express the area A of the triangle as a function of x. 9. A rectangle is inscribed in a circle of radius 2. See the figure. Let P = 1x, y2 be the point in quadrant I that is a vertex of the rectangle and is on the circle. y 2 y P (x, y) y x3 2 (0, y) (x, y) 2 x 2 2 2 x y 4 (0, 0) x 6. A right triangle has one vertex on the graph of y = 9 - x2, x 7 0, at 1x, y2, another at the origin, and the third on the positive x-axis at 1x, 02. Express the area A of the triangle as a function of x. 7. A rectangle has one corner in quadrant I on the graph of y = 16 - x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See the figure (top, right). (a) Express the area A of the rectangle as a function of x. (b) What is the domain of A? (c) Graph A = A1x2. For what value of x is A largest? (a) Express the area A of the rectangle as a function of x. (b) Express the perimeter p of the rectangle as a function of x. (c) Graph A = A1x2. For what value of x is A largest? (d) Graph p = p1x2. For what value of x is p largest? 10. A circle of radius r is inscribed in a square. See the figure. r (a) Express the area A of the square as a function of the radius r of the circle. (b) Express the perimeter p of the square as a function of r. 264 CHAPTER 3 Functions and Their Graphs 11. Geometry A wire 10 meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. See the figure. 40 miles per hour (see the figure). Build a model that expresses the distance d between the cars as a function of the time t. [Hint: At t = 0, the cars leave the intersection.] x 4x N W 10 m 10 4x E S (a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the square. (b) What is the domain of A ? (c) Graph A = A1x2. For what value of x is A smallest? 12. Geometry A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped as a circle. (a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle. (b) What is the domain of A ? (c) Graph A = A1x2. For what value of x is A smallest? 13. Geometry A wire of length x is bent into the shape of a circle. (a) Express the circumference C of the circle as a function of x. (b) Express the area A of the circle as a function of x. 14. Geometry A wire of length x is bent into the shape of a square. (a) Express the perimeter p of the square as a function of x. (b) Express the area A of the square as a function of x. 15. Geometry A semicircle of radius r is inscribed in a rectangle so that the diameter of the semicircle is the length of the rectangle. See the figure. d 19. Uniform Motion Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 miles per hour. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 miles per hour. (a) Build a model that expresses the distance d between the cars as a function of time t. [Hint: At t = 0, the cars are 2 miles south and 3 miles east of the intersection, respectively.] (b) Use a graphing utility to graph d = d1t2. For what value of t is d smallest? 20. Inscribing a Cylinder in a Sphere Inscribe a right circular cylinder of height h and radius r in a sphere of fixed radius R. See the illustration. Express the volume V of the cylinder as a function of h. [Hint: V = pr 2 h. Note also the right triangle.] r r (a) Express the area A of the rectangle as a function of the radius r of the semicircle. (b) Express the perimeter p of the rectangle as a function of r. 16. Geometry An equilateral triangle is inscribed in a circle of radius r. See the figure. Express the circumference C of the circle as a function of the length x of a side of the triangle. [Hint: First show that r 2 = h Sphere 21. Inscribing a Cylinder in a Cone Inscribe a right circular cylinder of height h and radius r in a cone of fixed radius R and fixed height H. See the illustration. Express the volume V of the cylinder as a function of r. x2 .] 3 x R x [Hint: V = pr 2 h. Note also the similar triangles.] r r x 17. Geometry An equilateral triangle is inscribed in a circle of radius r. See the figure in Problem 16. Express the area A within the circle, but outside the triangle, as a function of the length x of a side of the triangle. 18. Uniform Motion Two cars leave an intersection at the same time. One is headed south at a constant speed of 30 miles per hour, and the other is headed west at a constant speed of H h R Cone SECTION 3.6 Mathematical Models: Building Functions 22. Installing Cable TV MetroMedia Cable is asked to provide service to a customer whose house is located 2 miles from the road along which the cable is buried. The nearest connection box for the cable is located 5 miles down the road. See the figure. 265 24. Filling a Conical Tank Water is poured into a container in the shape of a right circular cone with radius 4 feet and height 16 feet. See the figure. Express the volume V of the water in the cone as a function of the height h of the water. [Hint: The volume V of a cone of radius r and height h is 1 V = pr 2 h.] 3 House Stream 4 2 mi Box 5 mi h (a) If the installation cost is $500 per mile along the road and $700 per mile off the road, build a model that expresses the total cost C of installation as a function of the distance x (in miles) from the connection box to the point where the cable installation turns off the road. Find the domain of C = C 1x2. (b) Compute the cost if x = 1 mile. (c) Compute the cost if x = 3 miles. (d) Graph the function C = C 1x2. Use TRACE to see how the cost C varies as x changes from 0 to 5. (e) What value of x results in the least cost? 23. Time Required to Go from an Island to a Town An island is 2 miles from the nearest point P on a straight shoreline. A town is 12 miles down the shore from P. See the illustration. d2 Town P 12 x 2 mi x r 16 x 12 mi d1 Island (a) If a person can row a boat at an average speed of 3 miles per hour and the same person can walk 5 miles per hour, build a model that expresses the time T that it takes to go from the island to town as a function of the distance x from P to where the person lands the boat. (b) What is the domain of T ? (c) How long will it take to travel from the island to town if the person lands the boat 4 miles from P? (d) How long will it take if the person lands the boat 8 miles from P? 25. Constructing an Open Box An open box with a square base is to be made from a square piece of cardboard 24 inches on a side by cutting out a square from each corner and turning up the sides. See the figure. x x x x 24 in. x x x x 24 in. (a) Express the volume V of the box as a function of the length x of the side of the square cut from each corner. (b) What is the volume if a 3-inch square is cut out? (c) What is the volume if a 10-inch square is cut out? (d) Graph V = V 1x2. For what value of x is V largest? 26. Constructing an Open Box An open box with a square base is required to have a volume of 10 cubic feet. (a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base. (b) How much material is required for a base 1 foot by 1 foot? (c) How much material is required for a base 2 feet by 2 feet? (d) Use a graphing utility to graph A = A1x2. For what value of x is A smallest? 266 CHAPTER 3 Functions and Their Graphs Retain Your Knowledge Problems 27–30 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 27. Solve: 2x - 3 - 5 = - 2 28. A 16-foot long Ford Fusion wants to pass a 50-foot truck traveling at 55 mi/h. How fast must the car travel to completely pass the truck in 5 seconds? 10 29. Find the slope of the line containing the points (3, - 2) and (1, 6). 30. Find the missing length x for the given pair of similar triangles. 4 x 14 Chapter Review Library of Functions Square function (p. 239) Identity function (p. 239) Constant function (p. 239) f 1x2 = b f 1x2 = x The graph is a horizontal line with y-intercept b. f 1x2 = x2 The graph is a line with slope 1 and y-intercept 0. y y 3 y f (x ) = b The graph is a parabola with intercept at 10, 02. ( – 2, 4) (2, 4) 4 (0,b) (1, 1) x (– 1, 1) 3 x (0, 0) –3 ( – 1, – 1) f 1x2 = x Square root function (p. 240) y y 4 2 (1, 1) 4 (1, 1) Cube root function (p. 240) f 1x2 = 1x 3 (0, 0) 4 x 1 4 x (0, 0) –4 Cube function (p. 240) (1, 1) 3 f 1x2 = 2x y 3 (1, 1) (4, 2) 5 x (0, 0) 3 ( 1–8 , 1–2 (1, 1) ) (2, 2 ) ( 1–8 , 1–2) 3 3 x (0, 0) 3 4 (1, 1) (2, 2 ) 3 Absolute value function (p. 240) Reciprocal function (p. 240) f 1x2 = f 1x2 = 0 x 0 1 x y y 2 3 2 (1, 1) 2 x (1, 1) 3 f 1x2 = int 1x2 y 4 (2, 2) (2, 2) (1, 1) Greatest integer function (p. 241) (0, 0) 2 (1, 1) 3 x 2 2 3 2 4 x Chapter Review 267 Things to Know Function (pp. 199–202) A relation between two sets so that each element x in the first set, the domain, has corresponding to it exactly one element y in the second set, the range. The range is the set of y-values of the function for the x-values in the domain. A function can also be characterized as a set of ordered pairs 1x, y2 in which no first element is paired with two different second elements. Function notation (pp. 202–205) y = f 1x2 f is a symbol for the function. x is the argument, or independent variable. y is the dependent variable. f 1x2 is the value of the function at x, or the image of x. A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f 1x2. Difference quotient of f (p. 205) f 1x + h2 - f 1x2 h h ≠ 0 Domain (pp. 206–208) If unspecified, the domain of a function f defined by an equation is the largest set of real numbers for which f 1x2 is a real number. Vertical-line test (p. 214) A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. Even function f (p. 224) Odd function f (p. 224) f 1 - x2 = f 1x2 for every x in the domain ( - x must also be in the domain). f 1 - x2 = - f 1x2 for every x in the domain ( - x must also be in the domain). Increasing function (p. 226) A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2. Decreasing function (p. 226) A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 7 f 1x2 2. Constant function (p. 226) A function f is constant on an open interval I if, for all choices of x in I, the values of f 1x2 are equal. Local maximum (p. 227) A function f , defined on some interval I, has a local maximum at c if there is an open interval in I containing c such that, for all x in this open interval, f 1x2 … f 1c2. Local minimum (p. 227) A function f , defined on some interval I, has a local minimum at c if there is an open interval in I containing c such that, for all x in this open interval, f 1x2 Ú f 1c2. Absolute maximum and Absolute minimum (p. 227) Let f denote a function defined on some interval I. If there is a number u in I for which f 1x2 … f 1u2 for all x in I, then f has an absolute maximum at u, and the number f 1u2 is the absolute maximum of f on I. If there is a number v in I for which f 1x2 Ú f 1v2 , for all x in I, then f has an absolute minimum at v, and the number f 1v2 is the absolute minimum of f on I. Average rate of change of a function (p. 230) The average rate of change of f from a to b is f 1b2 - f 1a2 ∆y = ∆x b - a a ≠ b Objectives Section You should be able to . . . Examples Review Exercises 3.1 1–5 6, 7 8 9, 10 1, 2 3–5, 39 15 6–11 11 1 12–14 27, 28 2–4 16(a)–(e), 17(a), 17(e), 17(g) 1 2 3 4 5 3.2 1 2 Determine whether a relation represents a function (p. 199) Find the value of a function (p. 202) Find the difference quotient of a function (p. 205) Find the domain of a function defined by an equation (p. 206) Form the sum, difference, product, and quotient of two functions (p. 208) Identify the graph of a function (p. 214) Obtain information from or about the graph of a function (p. 215) 268 CHAPTER 3 Functions and Their Graphs Section 3.3 1 2 3 4 5 6 7 3.4 1 2 3.5 1 2 3 3.6 1 You should be able to . . . Examples Review Exercises Determine even and odd functions from a graph (p. 223) Identify even and odd functions from an equation (p. 225) Use a graph to determine where a function is increasing, decreasing, or constant (p. 225) Use a graph to locate local maxima and local minima (p. 226) Use a graph to locate the absolute maximum and the absolute minimum (p. 227) Use a graphing utility to approximate local maxima and local minima and to determine where a function is increasing or decreasing (p. 229) Find the average rate of change of a function (p. 230) Graph the functions listed in the library of functions (p. 237) Graph piecewise-defined functions (p. 242) Graph functions using vertical and horizontal shifts (p. 248) Graph functions using compressions and stretches (p. 251) Graph functions using reflections about the x-axis and the y-axis (p. 253) Build and analyze functions (p. 260) 1 2 17(f) 18–21 3 4 17(b) 17(c) 5 17(d) 6 7, 8 1, 2 3, 4 1–5, 11–13 6–8, 12 22, 23, 40(d), 41(b) 24–26 29, 30 37, 38 16(f), 31, 33–36 16(g), 32, 36 9, 10, 11, 13 1–3 16(h), 32, 34, 36 40, 41 Review Exercises In Problems 1 and 2, determine whether each relation represents a function. For each function, state the domain and range. 1. 5 1 - 1, 02, 12, 32, 14, 02 6 2. 5 14, - 12, 12, 12, 14, 22 6 In Problems 3–5, find the following for each function: (a) f 122 (b) f 1 - 22 3x 3. f 1x2 = 2 x - 1 (c) f 1 - x2 (d) - f 1x2 (e) f 1x - 22 (f) f 12x2 5. f 1x2 = 4. f 1x2 = 2x2 - 4 In Problems 6–11, find the domain of each function. 6. f 1x2 = x x - 9 9. f 1x2 = x x + 2x - 3 7. f 1x2 = 22 - x 2 2 10. f(x) = In Problems 12–14, find f + g, f - g, f # g, and 12. f 1x2 = 2 - x; g1x2 = 3x + 1 8. g1x2 = 2x + 1 x2 - 4 11. g(x) = x2 - 4 x2 0x0 x x 2x + 8 f for each pair of functions. State the domain of each of these functions. g 13. f 1x2 = 3x2 + x + 1; g1x2 = 3x 15. Find the difference quotient of f 1x2 = - 2x2 + x + 1; that is, find 16. Consider the graph of the function f on the right. (a) Find the domain and the range of f. (b) List the intercepts. (c) Find f 1 - 22. (d) For what value of x does f 1x2 = - 3? (e) Solve f 1x2 7 0. (f) Graph y = f 1x - 32. 1 (g) Graph y = f a xb. 2 (h) Graph y = - f 1x2. 14. f 1x2 = f 1x + h2 - f 1x2 h x + 1 1 ; g1x2 = x - 1 x , h ≠ 0. y 4 5 (2, 1) (4, 3) (3, 3) (0, 0) 4 5 x Chapter Review 17. Use the graph of the function f shown to find: (a) The domain and the range of f . (b) The intervals on which f is increasing, decreasing, or constant. (c) The local minimum values and local maximum values. (d) The absolute maximum and absolute minimum. (e) Whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin. (f) Whether the function is even, odd, or neither. (g) The intercepts, if any. y 4 (3, 0) (22, 1) 26 (24,23) (23, 0) 269 (4, 3) 6 x (2, 21) 24 In Problems 18–21, determine (algebraically) whether the given function is even, odd, or neither. 18. f 1x2 = x3 - 4x 19. g1x2 = 4 + x2 1 + x4 21. f 1x2 = 20. G1x2 = 1 - x + x3 x 1 + x2 In Problems 22 and 23, use a graphing utility to graph each function over the indicated interval. Approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. 22. f 1x2 = 2x3 - 5x + 1 1 - 3, 32 24. Find the average rate of change of f 1x2 = 8x - x: (a) From 1 to 2 (b) From 0 to 1 (c) From 2 to 4 23. f 1x2 = 2x4 - 5x3 + 2x + 1 1 - 2, 32 2 In Problems 25 and 26, find the average rate of change from 2 to 3 for each function f. Be sure to simplify. 25. f 1x2 = 2 - 5x 26. f 1x2 = 3x - 4x2 In Problems 27 and 28, is the graph shown the graph of a function? 27. 28. y y x x In Problems 29 and 30, graph each function. Be sure to label at least three points. 29. f 1x2 = 0 x 0 30. f 1x2 = 1x In Problems 31–36, graph each function using the techniques of shifting, compressing or stretching, and reflections. Identify any intercepts of the graph. State the domain and, based on the graph, find the range. 31. F 1x2 = 0 x 0 - 4 32. g1x2 = - 2 0 x 0 33. h1x2 = 2x - 1 34. f 1x2 = 21 - x 35. h1x2 = 1x - 12 2 + 2 36. g1x2 = - 21x + 22 3 - 8 In Problems 37 and 38: (a) Find the domain of each function. (d) Based on the graph, find the range. 37. f1x2 = b 3x x + 1 (b) Locate any intercepts. (e) Is f continuous on its domain? if - 2 6 x … 1 if x 7 1 39. A function f is defined by f 1x2 = x 38. f 1x2 = c 1 3x (c) Graph each function. if - 4 … x 6 0 if x = 0 if x 7 0 Ax + 5 6x - 2 If f 112 = 4, find A. 40. Constructing a Closed Box A closed box with a square base is required to have a volume of 10 cubic feet. (a) Build a model that expresses the amount A of material used to make such a box as a function of the length x of a side of the square base. (b) How much material is required for a base 1 foot by 1 foot? (c) How much material is required for a base 2 feet by 2 feet? (d) Graph A = A1x2. For what value of x is A smallest? 41. Area of a Rectangle A rectangle has one vertex in quadrant I on the graph of y = 10 - x2, another at the origin, one on the positive x-axis, and one on the positive y-axis. (a) Express the area A of the rectangle as a function of x. (b) Find the largest area A that can be enclosed by the rectangle. 270 CHAPTER 3 Functions and Their Graphs The Chapter Test Prep Videos are step-by-step solutions available in Channel. Flip back to the Resources , or on this text’s for Success page for a link to this text’s YouTube channel. Chapter Test 1. Determine whether each relation represents a function. For each function, state the domain and the range. (a) 5 12, 52, 14, 62, 16, 72, 18, 82 6 (b) 5 11, 32, 14, - 22, 1 - 3, 52, 11, 72 6 (c) 7. Consider the function g1x2 = b y 6 (a) (b) (c) (d) 4 2 x 4 2 2 4 4 y 4 2 x 2 2 4 2 3. g1x2 = x + 2 0x + 20 x - 4 x2 + 5x - 36 5. Consider the graph of the function f below. 4. h1x2 = 4 (1, 3) (0, 2) (2, 0) 4 (a) (b) (c) (d) (e) x 4 2 (5, 3) 4 8. For the function f 1x2 = 3x2 - 2x + 4, find the average rate of change of f from 3 to 4. 11. The variable interest rate on a student loan changes each July 1 based on the bank prime loan rate. For the years 1992–2007, this rate can be approximated by the model (5, 2) (3, 3) Find the domain and the range of f. List the intercepts. Find f 112. For what value(s) of x does f 1x2 = - 3? Solve f 1x2 6 0. where x is the number of years since 1992 and r is the interest rate as a percent. (a) Use a graphing utility to estimate the highest rate during this time period. During which year was the interest rate the highest? (b) Use the model to estimate the rate in 2010. Does this value seem reasonable? Source: U.S. Federal Reserve 12. A community skating rink is in the shape of a rectangle with semicircles attached at the ends. The length of the rectangle is 20 feet less than twice the width. The thickness of the ice is 0.75 inch. (a) Build a model that expresses the ice volume, V, as a function of the width, x. (b) How much ice is in the rink if the width is 90 feet? y (2, 0) Graph the function. List the intercepts. Find g1 - 52. Find g122. r 1x2 = - 0.115x2 + 1.183x + 5.623, In Problems 2–4, find the domain of each function and evaluate each function at x = - 1. 2. f 1x2 = 24 - 5x if x 6 - 1 if x Ú - 1 10. Graph each function using the techniques of shifting, compressing or stretching, and reflecting. Start with the graph of the basic function and show all stages. (a) h1x2 = - 21x + 12 3 + 3 (b) g1x2 = 0 x + 4 0 + 2 6 4 2x + 1 x - 4 9. For the functions f 1x2 = 2x2 + 1 and g1x2 = 3x - 2, find the following and simplify. (a) 1f - g2 1x2 (b) 1f # g2 1x2 (c) f 1x + h2 - f 1x2 2 (d) 6. Use a graphing utility to graph the function f 1x2 = - x4 + 2x3 + 4x2 - 2 on the interval 1 - 5, 52. Then approximate any local maximum values and local minimum values rounded to two decimal places. Determine where the function is increasing and where it is decreasing. Chapter Projects 271 Cumulative Review In Problems 1–6, find the real solutions of each equation. In Problems 11–14, graph each equation. 11. 3x - 2y = 12 12. x = y2 1. 3x - 8 = 10 2. 3x - x = 0 3. x - 8x - 9 = 0 4. 6x - 5x + 1 = 0 5. 0 2x + 3 0 = 4 15. For the equation 3x2 - 4y = 12, find the intercepts and check for symmetry. 6. 22x + 3 = 2 16. Find the slope–intercept form of the equation of the line containing the points 1 - 2, 42 and 16, 82 . 2 2 2 13. x2 + 1y - 32 2 = 16 14. y = 2x In Problems 7–9, solve each inequality. Graph the solution set. 7. 2 - 3x 7 6 8. 0 2x - 5 0 6 3 9. 0 4x + 1 0 Ú 7 10. (a) Find the distance from P1 = 1 - 2, - 32 to P2 = 13, - 52. (b) What is the midpoint of the line segment from P1 to P2? (c) What is the slope of the line containing the points P1 and P2? In Problems 17–19, graph each function. 17. f 1x2 = 1x + 22 2 - 3 18. f 1x2 = 1 x 19. f 1x2 = e 2 - x 0x0 if x … 2 if x 7 2 Chapter Projects that include unlimited talk and text. The monthly cost is primarily determined by the amount of data used and the number of devices. 1. Suppose you expect to use 10 gigabytes of data for a single smartphone. What would be the monthly cost of each plan you are considering? 2. Suppose you expect to use 30 gigabytes of data and want a personal hotspot, but you still have only a single smartphone. What would be the monthly cost of each plan you are considering? 3. Suppose you expect to use 20 gigabytes of data with three smartphones sharing the data. What would be the monthly cost of each plan you are considering? 4. Suppose you expect to use 20 gigabytes of data with a single smartphone and a personal hotspot. What would be the monthly cost of each plan you are considering? 5. Build a model that describes the monthly cost C, in dollars, as a function of the number of data gigabytes used, g, assuming a single smartphone and a personal hotspot for each plan you are considering. 6. Graph each function from Problem 5. Internet-based Project I. Choosing a Wireless Data Plan Collect information from your family, friends, or consumer agencies such as Consumer Reports. Then decide on a cellular provider, choosing the company that you feel offers the best service. Once you have selected a service provider, research the various types of individual plans offered by the company by visiting the provider’s website. Many providers offer family plans 7. Based on your particular usage, which plan is best for you? 8. Now, develop an Excel spreadsheet to analyze the various plans you are considering. Suppose you want a family plan with unlimited talk and text that offers 10 gigabytes of shared data and costs $100 per month. Additional gigabytes of data cost $15 per gigabyte, extra phones can be added to the plan for $15 each per month, and each hotspot costs $20 per month. Because wireless 272 CHAPTER 3 Functions and Their Graphs data plans have a cost structure based on piecewise-defined functions, we need an “if/then” statement within Excel to analyze the cost of the plan. Use the accompanying Excel spreadsheet as a guide in developing your spreadsheet. Enter into your spreadsheet a variety of possible amounts of data and various numbers of additional phones and hotspots. A 1 2 Monthly fee 3 Allotted data per month (GB) 4 Data used (GB) 5 6 7 8 9 10 B D $100 10 Cost per additional GB of data 12 $15 Monthly cost of hotspot Number of hotspots Monthly cost of additional phone Number of additional phones $20 1 $15 2 11 12 Cost of data 13 Cost of additional devices/hotspots 14 15 Total Cost 16 C =IF(B4<B3,B2,B2+B5*(B4-B3)) =B8*B7+B10*B9 =B12+B13 9. Write a paragraph supporting the choice in plans that best meets your needs. 10. How are “if/then” loops similar to a piecewise-defined function? Citation: Excel © 2013 Microsoft Corporation. Used with permission from Microsoft. The following projects are available on the Instructor’s Resource Center (IRC). II. Project at Motorola: Wireless Internet Service Use functions and their graphs to analyze the total cost of various wireless Internet service plans. III. Cost of Cable When government regulations and customer preference influence the path of a new cable line, the Pythagorean Theorem can be used to assess the cost of installation. IV. Oil Spill Functions are used to analyze the size and spread of an oil spill from a leaking tanker. 4 Linear and Quadratic Functions The Beta of a Stock Investing in the stock market can be rewarding and fun, but how does one go about selecting which stocks to purchase? Financial investment firms hire thousands of analysts who track individual stocks (equities) and assess the value of the underlying company. One measure the analysts consider is the beta of the stock. Beta measures the risk of an individual company’s equity relative to that of a market basket of stocks, such as the Standard & Poor’s 500. But how is beta computed? —See the Internet-based Chapter Project I— A Look Back Up to now, our discussion has focused on graphs of equations and functions. We learned how to graph equations using the point-plotting method, intercepts, and the tests for symmetry. In addition, we learned what a function is and how to identify whether a relation represents a function. We also discussed properties of functions, such as domain/range, increasing/decreasing, even/odd, and average rate of change. A Look Ahead Going forward, we will look at classes of functions. This chapter focuses on linear and quadratic functions, their properties, and their applications. Outline 4.1 4.2 4.3 4.4 4.5 Properties of Linear Functions and Linear Models Building Linear Models from Data Quadratic Functions and Their Properties Build Quadratic Models from Verbal Descriptions and from Data Inequalities Involving Quadratic Functions Chapter Review Chapter Test Cumulative Review Chapter Projects 273 273 274 CHAPTER 4 Linear and Quadratic Functions 4.1 Properties of Linear Functions and Linear Models PREPARING FOR THIS SECTION Before getting started, review the following: r Functions (Section 3.1, pp. 199–208) r The Graph of a Function (Section 3.2, pp. 214–217) r Properties of Functions (Section 3.3, pp. 223–231) r Lines (Section 2.3, pp. 167–175) r Graphs of Equations in Two Variables; Intercepts; Symmetry (Section 2.2, pp. 157–164) r Linear Equations (Section 1.1, pp. 82–87) Now Work the ‘Are You Prepared?’ problems on page 280. OBJECTIVES 1 Graph Linear Functions (p. 274) 2 Use Average Rate of Change to Identify Linear Functions (p. 274) 3 Determine Whether a Linear Function Is Increasing, Decreasing, or Constant (p. 277) 4 Build Linear Models from Verbal Descriptions (p. 278) 1 Graph Linear Functions In Section 2.3 we discussed lines. In particular, for nonvertical lines we developed the slope–intercept form of the equation of a line y = mx + b. When the slope–intercept form of a line is written using function notation, the result is a linear function. DEFINITION A linear function is a function of the form f1x2 = mx + b The graph of a linear function is a line with slope m and y-intercept b. Its domain is the set of all real numbers. Functions that are not linear are said to be nonlinear. EX AMPLE 1 Solution y (0, 7) Δx 1 1 3 This is a linear function with slope m = - 3 and y-intercept b = 7. To graph this function, plot the point 10, 72, the y-intercept, and use the slope to find an additional point by moving right 1 unit and down 3 units. See Figure 1. The domain and the range of f are each the set of all real numbers. Alternatively, an additional point could have been found by evaluating the function at some x ≠ 0. For x = 1, f 112 = - 3112 + 7 = 4 and the point 11, 42 lies on the graph. (1, 4) 3 1 Graph the linear function f 1x2 = - 3x + 7. What are the domain and the range of f ? r Δy 3 5 Graphing a Linear Function 5 x Figure 1 f 1x2 = - 3x + 7 Now Work PROBLEMS 13(a) AND (b) 2 Use Average Rate of Change to Identify Linear Functions Look at Table 1, which shows certain values of the independent variable x and corresponding values of the dependent variable y for the function f 1x2 = - 3x + 7. Notice that as the value of the independent variable, x, increases by 1, the value of the dependent variable y decreases by 3. That is, the average rate of change of y with respect to x is a constant, - 3. SECTION 4.1 Properties of Linear Functions and Linear Models Table 1 x y = f(x) = −3x + 7 -2 13 Average Rate of Change = 275 𝚫y 𝚫x 10 - 13 -3 = = -3 - 1 - 1 - 22 1 -1 10 -3 7 - 10 = = -3 0 - 1 - 12 1 0 7 -3 1 4 2 1 3 -2 -3 -3 It is not a coincidence that the average rate of change of the linear function ∆y f1x2 = - 3x + 7 is the slope of the linear function. That is, = m = - 3. The ∆x following theorem states this fact. THEOREM Average Rate of Change of a Linear Function Linear functions have a constant average rate of change. That is, the average rate of change of a linear function f 1x2 = mx + b is ∆y = m ∆x Proof The average rate of change of f1x2 = mx + b from x1 to x2, x1 ≠ x2, is f1x2 2 - f1x1 2 1mx2 + b2 - 1mx1 + b2 ∆y = = x2 - x1 x2 - x1 ∆x = m1x2 - x1 2 mx2 - mx1 = = m x2 - x1 x2 - x1 ■ Based on the theorem just proved, the average rate of change of the function 2 2 g1x2 = - x + 5 is - . 5 5 Now Work PROBLEM 13(C) As it turns out, only linear functions have a constant average rate of change. Because of this, the average rate of change can be used to determine whether a function is linear. This is especially useful if the function is defined by a data set. EXAMPL E 2 Using the Average Rate of Change to Identify Linear Functions (a) A strain of E. coli known as Beu 397-recA441 is placed into a Petri dish at 30° Celsius and allowed to grow. The data shown in Table 2 on the next page are collected. The population is measured in grams and the time in hours. Plot the ordered pairs 1x, y2 in the Cartesian plane, and use the average rate of change to determine whether the function is linear. 276 CHAPTER 4 Linear and Quadratic Functions (b) The data in Table 3 represent the maximum number of heartbeats that a healthy individual of different ages should have during a 15-second interval of time while exercising. Plot the ordered pairs 1x, y2 in the Cartesian plane, and use the average rate of change to determine whether the function is linear. Table 3 Table 2 Maximum Number of Heartbeats, y Time (hours), x Population (grams), y (x, y) Age, x 0 0.09 (0, 0.09) 20 50 (20, 50) 1 0.12 (1, 0.12) 30 47.5 (30, 47.5) 2 0.16 (2, 0.16) 40 45 (40, 45) 3 0.22 (3, 0.22) 50 42.5 (50, 42.5) 4 0.29 (4, 0.29) 60 40 (60, 40) 5 0.39 (5, 0.39) 70 37.5 (70, 37.5) (x, y ) Source: American Heart Association Solution Compute the average rate of change of each function. If the average rate of change is constant, the function is linear. If the average rate of change is not constant, the function is nonlinear. (a) Figure 2 shows the points listed in Table 2 plotted in the Cartesian plane. Note that it is impossible to draw a straight line that contains all the points. Table 4 displays the average rate of change of the population. Population (grams), y y Table 4 0.4 Time (hours), x Population (grams), y 0 0.09 𝚫y 𝚫x 0.12 - 0.09 = 0.03 1 - 0 0.3 0.2 1 0.12 2 0.16 0.04 0.1 0 1 2 3 4 5 x 0.06 Time (hours), x Figure 2 Average Rate of Change = 3 0.22 0.07 4 0.29 0.10 5 0.39 Because the average rate of change is not constant, the function is not linear. In fact, because the average rate of change is increasing as the value of the independent variable increases, the function is increasing at an increasing rate. So not only is the population increasing over time, but it is also growing more rapidly as time passes. (b) Figure 3 shows the points listed in Table 3 plotted in the Cartesian plane. Note that the data in Figure 3 lie on a straight line. Table 5 displays the average rate of change of the maximum number of heartbeats. The average rate of change of the heartbeat data is constant, - 0.25 beat per year, so the function is linear. SECTION 4.1 Properties of Linear Functions and Linear Models Table 5 y Heartbeats 50 Age, x Maximum Number of Heartbeats, y 20 50 Average Rate of Change = 277 𝚫y 𝚫x 47.5 - 50 = - 0.25 30 - 20 45 30 47.5 40 40 45 - 0.25 - 0.25 20 30 40 50 60 70 x 50 42.5 Age - 0.25 60 Figure 3 40 - 0.25 70 37.5 r Now Work PROBLEM 21 3 Determine Whether a Linear Function Is Increasing, Decreasing, or Constant Look back at the Seeing the Concept on page 169. When the slope m of a linear function is positive (m 7 0), the line slants upward from left to right. When the slope m of a linear function is negative (m 6 0), the line slants downward from left to right. When the slope m of a linear function is zero (m = 0), the line is horizontal. THEOREM Increasing, Decreasing, and Constant Linear Functions A linear function f1x2 = mx + b is increasing over its domain if its slope, m, is positive. It is decreasing over its domain if its slope, m, is negative. It is constant over its domain if its slope, m, is zero. EXAMPL E 3 Determining Whether a Linear Function Is Increasing, Decreasing, or Constant Determine whether the following linear functions are increasing, decreasing, or constant. (a) f1x2 = 5x - 2 3 (c) s 1t2 = t - 4 4 Solution (b) g1x2 = - 2x + 8 (d) h 1z2 = 7 (a) For the linear function f1x2 = 5x - 2, the slope is 5, which is positive. The function f is increasing on the interval 1 - q , q 2. (b) For the linear function g1x2 = - 2x + 8, the slope is - 2, which is negative. The function g is decreasing on the interval 1 - q , q 2. 3 3 (c) For the linear function s 1t2 = t - 4, the slope is , which is positive. The 4 4 function s is increasing on the interval 1 - q , q 2. (d) The linear function h can be written as h 1z2 = 0z + 7. Because the slope is 0, the function h is constant on the interval 1 - q , q 2. r Now Work PROBLEM 13 (d) 278 CHAPTER 4 Linear and Quadratic Functions 4 Build Linear Models from Verbal Descriptions When the average rate of change of a function is constant, a linear function can model the relation between the two variables. For example, if a recycling company pays $0.52 per pound for aluminum cans, then the relation between the price paid p and the pounds recycled x can be modeled as the linear function p1x2 = 0.52x, 0.52 dollar with slope m = . 1 pound Modeling with a Linear Function If the average rate of change of a function is a constant m, a linear function f can be used to model the relation between the two variables as follows: f1x2 = mx + b where b is the value of f at 0; that is, b = f102. EX AMPLE 4 Straight-line Depreciation Book value is the value of an asset that a company uses to create its balance sheet. Some companies depreciate assets using straight-line depreciation so that the value of the asset declines by a fixed amount each year. The amount of the decline depends on the useful life that the company assigns to the asset. Suppose a company just purchased a fleet of new cars for its sales force at a cost of $31,500 per car. The company chooses to depreciate each vehicle using the straight-line method over 7 years. This +31,500 means that each car will depreciate by = +4500 per year. 7 (a) Write a linear function that expresses the book value V of each car as a function of its age, x, in years. (b) Graph the linear function. (c) What is the book value of each car after 3 years? (d) Interpret the slope. (e) When will the book value of each car be $9000? [Hint: Solve the equation V 1x2 = 9000.] Solution v 31,500 V 1x2 = - 4500x + 31,500 27,000 Book value ($) (a) If we let V 1x2 represent the value of each car after x years, then V 102 represents the original value of each car, so V 102 = +31,500. The y-intercept of the linear function is $31,500. Because each car depreciates by $4500 per year, the slope of the linear function is - 4500. The linear function that represents the book value V of each car after x years is 22,500 (b) Figure 4 shows the graph of V. (c) The book value of each car after 3 years is 18,000 13,500 V 132 = - 4500132 + 31,500 = +18,000 9000 4500 1 2 3 4 5 6 7 x Age of vehicle (years) Figure 4 V 1 x2 = - 4500x + 31, 500 (d) Since the slope of V 1x2 = - 4500x + 31,500 is - 4500, the average rate of change of the book value is - $4500/year. So for each additional year that passes, the book value of the car decreases by $4500. SECTION 4.1 Properties of Linear Functions and Linear Models 279 (e) To find when the book value will be $9000, solve the equation V 1x2 = 9000 - 4500x + 31,500 = 9000 - 4500x = - 22,500 - 22,500 x = = 5 - 4500 Subtract 31,500 from each side. Divide by - 4500. The car will have a book value of $9000 when it is 5 years old. Now Work EXAMPL E 5 PROBLEM r 45 Supply and Demand The quantity supplied of a good is the amount of a product that a company is willing to make available for sale at a given price. The quantity demanded of a good is the amount of a product that consumers are willing to purchase at a given price. Suppose that the quantity supplied, S, and the quantity demanded, D, of cellular telephones each month are given by the following functions: S 1p2 = 60p - 900 D1p2 = - 15p + 2850 where p is the price (in dollars) of the telephone. (a) The equilibrium price of a product is defined as the price at which quantity supplied equals quantity demanded. That is, the equilibrium price is the price at which S 1p2 = D1p2. Find the equilibrium price of cellular telephones. What is the equilibrium quantity, the amount demanded (or supplied) at the equilibrium price? (b) Determine the prices for which quantity supplied is greater than quantity demanded. That is, solve the inequality S 1p2 7 D1p2. (c) Graph S = S 1p2 and D = D1p2, and label the equilibrium point, the point of intersection of S and D. Solution (a) To find the equilibrium price, solve the equation S 1p2 = D1p2. 60p - 900 = - 15p + 2850 60p = - 15p + 3750 75p = 3750 p = 50 S1p2 = 60p - 900; D (p) = - 15p + 2850 Add 900 to each side. Add 15p to each side. Divide each side by 75. The equilibrium price is $50 per cellular phone. To find the equilibrium quantity, evaluate either S 1p2 or D 1p2 at p = 50. S 1502 = 601502 - 900 = 2100 The equilibrium quantity is 2100 cellular phones. At a price of $50 per phone, the company will produce and sell 2100 phones each month and have no shortages or excess inventory. (b) The inequality S 1p2 7 D1p2 is 60p - 900 60p 75p p 7 7 7 7 - 15p + 2850 - 15p + 3750 3750 50 S1p2 7 D1p2 Add 900 to each side. Add 15p to each side. Divide each side by 75. If the company charges more than $50 per phone, quantity supplied will exceed quantity demanded. In this case the company will have excess phones in inventory. 280 CHAPTER 4 Linear and Quadratic Functions (c) Figure 5 shows the graphs of S = S 1p2 and D = D1p2 with the equilibrium point labeled. Quantity supplied, Quantity demanded S, D S 5 S(p) 3000 (0, 2850) Equilibrium point (50, 2100) 2000 D 5 D(p) 1000 (15, 0) 50 100 p Price ($) r Figure 5 Supply and demand functions Now Work PROBLEM 39 4.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Graph y = x2 - 1. (pp. 157–164) 4. Solve: 60x - 900 = - 15x + 2850. (pp. 82–87) 2. Find the slope of the line joining the points 12, 52 and 5. If f 1x2 = x2 - 4, find f 1 - 22. (pp. 199–208) 6. True or False The graph of the function f 1x2 = x2 is increasing on the interval 10, q 2. (pp. 214–217) 1 - 1, 32. (pp. 167–175) 3. Find the average rate of change of f 1x2 = 3x2 - 2, from 2 to 4. (pp. 223–231) Concepts and Vocabulary 7. For the graph of the linear function f 1x2 = mx + b, m is the and b is the . 8. If the slope m of the graph of a linear function is the function is increasing over its domain. 11. What is the only type of function that has a constant average rate of change? (a) linear function (b) quadratic function (c) step function (d) absolute value function , 12. A car has 12,500 miles on its odometer. Say the car is driven an average of 40 miles per day. Choose the model that expresses the number of miles N that will be on its odometer after x days. (a) N(x) = - 40x + 12,500 (b) N(x) = 40x - 12,500 (c) N(x) = 12,500x + 40 (d) N(x) = 40x + 12,500 9. True or False The slope of a nonvertical line is the average rate of change of the linear function. 10. True or False The average rate of change of f 1x2 = 2x + 8 is 8. Skill Building In Problems 13–20, a linear function is given. (a) Determine the slope and y-intercept of each function. (b) Use the slope and y-intercept to graph the linear function. (c) Determine the average rate of change of each function. (d) Determine whether the linear function is increasing, decreasing, or constant. 13. f 1x2 = 2x + 3 14. g1x2 = 5x - 4 15. h1x2 = - 3x + 4 16. p1x2 = - x + 6 1 17. f 1x2 = x - 3 4 2 18. h1x2 = - x + 4 3 19. F 1x2 = 4 20. G1x2 = - 2 In Problems 21–28, determine whether the given function is linear or nonlinear. If it is linear, determine the slope. 21. x y = f (x) -2 -1 22. x y = f (x) 4 -2 1 -1 23. x y = f (x) 1/4 -2 1/2 -1 24. x y = f (x) -8 -2 -4 -3 -1 0 0 -2 0 1 0 0 0 4 1 -5 1 2 1 1 1 8 2 -8 2 4 2 0 2 12 SECTION 4.1 Properties of Linear Functions and Linear Models 25. 26. x y = f (x) -2 - 26 -2 -1 -4 -1 x 27. x y = f (x) -4 -2 - 3.5 -1 y = f (x) 28. x y = f (x) 8 -2 0 8 -1 1 0 2 0 -3 0 8 0 4 1 -2 1 - 2.5 1 8 1 9 2 - 10 2 -2 2 8 2 16 281 Applications and Extensions 29. Suppose that f 1x2 = 4x - 1 and g1x2 = - 2x + 5. (a) Solve f 1x2 = 0. (b) Solve f 1x2 7 0. (c) Solve f 1x2 = g1x2. (d) Solve f 1x2 … g1x2. (e) Graph y = f 1x2 and y = g1x2 and label the point that represents the solution to the equation f 1x2 = g1x2. 30. Suppose that f 1x2 = 3x + 5 and g1x2 = - 2x + 15. (a) Solve f 1x2 = 0. (b) Solve f 1x2 6 0. (c) Solve f 1x2 = g1x2. (d) Solve f 1x2 Ú g1x2. (e) Graph y = f 1x2 and y = g1x2 and label the point that represents the solution to the equation f 1x2 = g1x2. 31. In parts (a)–(f), use the following figure. y y f (x) 34. In parts (a) and (b), use the following figure. y y f (x) (2, 5) x y g(x) (a) Solve the equation: f 1x2 = g1x2. (b) Solve the inequality: f 1x2 … g1x2. 35. In parts (a) and (b), use the following figure. y f (x ) y (88, 80) (0, 12) (40, 50) y h (x ) x (40, 0) (a) Solve f 1x2 = 50. (c) Solve f 1x2 = 0. (e) Solve f 1x2 … 80. (b) Solve f 1x2 = 80. (d) Solve f 1x2 7 50. (f) Solve 0 6 f 1x2 6 80. 32. In parts (a)–(f), use the following figure. y g(x ) (5, 12) y x (6,5) (0,5) y g (x ) (a) Solve the equation: f 1x2 = g1x2. (b) Solve the inequality: g1x2 … f 1x2 6 h1x2. 36. In parts (a) and (b), use the following figure. y (15, 60) (0, 7) y h (x) (4, 7) (5, 20) x (15, 0) x (a) Solve g1x2 = 20. (c) Solve g1x2 = 0. (e) Solve g1x2 … 60. (b) Solve g1x2 = 60. (d) Solve g1x2 7 20. (f) Solve 0 6 g1x2 6 60. 33. In parts (a) and (b), use the following figure. y f(x) y y g (x ) (4, 6) x (a) Solve the equation: f 1x2 = g1x2. (b) Solve the inequality: f 1x2 7 g1x2. (0, 8) (7,8) y g (x) y f (x) (a) Solve the equation: f 1x2 = g1x2. (b) Solve the inequality: g1x2 6 f 1x2 … h1x2. 37. Truck Rentals The cost C, in dollars, of a one-day truck rental is modeled by the function C 1x2 = 0.35x + 45, where x is the number of miles driven. (a) What is the cost if you drive x = 40 miles? (b) If the cost of renting the truck is $108, how many miles did you drive? (c) Suppose that you want the cost to be no more than $150. What is the maximum number of miles that you can drive? (d) What is the implied domain of C? 282 CHAPTER 4 Linear and Quadratic Functions 38. Phone Charges The monthly cost C, in dollars, for calls from the United States to Germany on a certain phone plan is modeled by the function C 1x2 = 0.26x + 5, where x is the number of minutes used. (a) What is the cost if you talk on the phone for x = 50 minutes? (b) Suppose that your monthly bill is $21.64. How many minutes did you use the phone? (c) Suppose that you budget yourself $50 per month for the phone. What is the maximum number of minutes that you can talk? (d) What is the implied domain of C if there are 30 days in the month? 39. Supply and Demand Suppose that the quantity supplied S and the quantity demanded D of T-shirts at a concert are given by the following functions: S1p2 = - 600 + 50p D1p2 = 1200 - 25p where p is the price of a T-shirt. (a) Find the equilibrium price for T-shirts at this concert. What is the equilibrium quantity? (b) Determine the prices for which quantity demanded is greater than quantity supplied. (c) What do you think will eventually happen to the price of T-shirts if quantity demanded is greater than quantity supplied? 40. Supply and Demand Suppose that the quantity supplied S and the quantity demanded D of hot dogs at a baseball game are given by the following functions: S1p2 = - 2000 + 3000p D1p2 = 10,000 - 1000p where p is the price of a hot dog. (a) Find the equilibrium price for hot dogs at the baseball game. What is the equilibrium quantity? (b) Determine the prices for which quantity demanded is less than quantity supplied. (c) What do you think will eventually happen to the price of hot dogs if quantity demanded is less than quantity supplied? 41. Taxes The function T 1x2 = 0.151x - 90752 + 907.50 represents the tax bill T of a single person whose adjusted gross income is x dollars for income between $9075 and $36,900, inclusive, in 2014. Source: Internal Revenue Service (a) What is the domain of this linear function? (b) What is a single filer’s tax bill if adjusted gross income is $20,000? (c) Which variable is independent and which is dependent? (d) Graph the linear function over the domain specified in part (a). (e) What is a single filer’s adjusted gross income if the tax bill is $3671.25? 42. Luxury Tax In 2011, major league baseball signed a labor agreement with the players. In this agreement, any team whose payroll exceeded $189 million in 2014 had to pay a luxury tax of 50%. The linear function T 1p2 = 0.501p - 1892 describes the luxury tax T for a team whose payroll was p (in millions of dollars). Source: Major League Baseball (a) What is the implied domain of this linear function? (b) What was the luxury tax for the New York Yankees, whose 2014 payroll was $203.4 million? (c) Graph the linear function. (d) What is the payroll of a team that pays a luxury tax of $15.7 million? The point at which a company’s profits equal zero is called the company’s break-even point. For Problems 43 and 44, let R represent a company’s revenue, let C represent the company’s costs, and let x represent the number of units produced and sold each day. (a) Find the firm’s break-even point; that is, find x so that R = C. (b) Find the values of x such that R 1x2 7 C 1x2. This represents the number of units that the company must sell to earn a profit. 43. R 1x2 = 8x C 1x2 = 4.5x + 17,500 44. R 1x2 = 12x C 1x2 = 10x + 15,000 45. Straight-line Depreciation Suppose that a company has just purchased a new computer for $3000. The company chooses to depreciate the computer using the straight-line method over 3 years. (a) Write a linear model that expresses the book value V of the computer as a function of its age x. (b) What is the implied domain of the function found in part (a)? (c) Graph the linear function. (d) What is the book value of the computer after 2 years? (e) When will the computer have a book value of $2000? 46. Straight-line Depreciation Suppose that a company has just purchased a new machine for its manufacturing facility for $120,000. The company chooses to depreciate the machine using the straight-line method over 10 years. (a) Write a linear model that expresses the book value V of the machine as a function of its age x. (b) What is the implied domain of the function found in part (a)? (c) Graph the linear function. (d) What is the book value of the machine after 4 years? (e) When will the machine have a book value of $72,000? 47. Cost Function The simplest cost function is the linear cost function, C 1x2 = mx + b, where the y-intercept b represents the fixed costs of operating a business and the slope m represents the cost of each item produced. Suppose that a small bicycle manufacturer has daily fixed costs of $1800, and each bicycle costs $90 to manufacture. (a) Write a linear model that expresses the cost C of manufacturing x bicycles in a day. (b) Graph the model. (c) What is the cost of manufacturing 14 bicycles in a day? (d) How many bicycles could be manufactured for $3780? 48. Cost Function Refer to Problem 47. Suppose that the landlord of the building increases the bicycle manufacturer’s rent by $100 per month. (a) Assuming that the manufacturer is open for business 20 days per month, what are the new daily fixed costs? (b) Write a linear model that expresses the cost C of manufacturing x bicycles in a day with the higher rent. (c) Graph the model. (d) What is the cost of manufacturing 14 bicycles in a day? (e) How many bicycles can be manufactured for $3780? 49. Truck Rentals A truck rental company rents a truck for one day by charging $31.95 plus $0.89 per mile. (a) Write a linear model that relates the cost C, in dollars, of renting the truck to the number x of miles driven. SECTION 4.1 Properties of Linear Functions and Linear Models (b) What is the cost of renting the truck if the truck is driven 110 miles? 230 miles? 50. International Call Plan A cell phone company offers an international plan by charging $30 for the first 80 minutes, plus $0.50 for each minute over 80. 283 (a) Write a linear model that relates the cost C, in dollars, of talking x minutes, assuming x Ú 80. (b) What is the cost of talking 105 minutes? 120 minutes? Mixed Practice 51. Building a Linear Model from Data How many songs can an iPod hold? The following data represent the memory m and the number of songs, n. Memory, m (gigabytes) Number of Songs, n 8 16 32 64 1750 3500 7000 14,000 (a) Plot the ordered pairs 1m, n2 in a Cartesian plane. (b) Show that the number of songs n is a linear function of the memory m. (c) Determine the linear function that describes the relation between m and n. (d) What is the implied domain of the linear function? (e) Graph the linear function in the Cartesian plane drawn in part (a). (f) Interpret the slope. 52. Building a Linear Model from Data The following data represent the various combinations of soda and hot dogs that Yolanda can buy at a baseball game with $60. Soda, s Hot Dogs, h 20 15 10 5 0 3 6 9 (a) Plot the ordered pairs 1s, h2 in a Cartesian plane. (b) Show that the number of hot dogs purchased h is a linear function of the number of sodas purchased s. (c) Determine the linear function that describes the relation between s and h. (d) What is the implied domain of the linear function? (e) Graph the linear function in the Cartesian plane drawn in part (a). (f) Interpret the slope. (g) Interpret the values of the intercepts. Explaining Concepts: Discussion and Writing 53. Which of the following functions might have the graph shown? (More than one answer is possible.) (a) f 1x2 = 2x - 7 y (b) g1x2 = - 3x + 4 (c) H1x2 = 5 (d) F 1x2 = 3x + 4 1 x (e) G1x2 = x + 2 2 54. Which of the following functions might have the graph shown? (More than one answer is possible.) (a) f 1x2 = 3x + 1 y (b) g1x2 = - 2x + 3 (c) H1x2 = 3 (d) F 1x2 = - 4x - 1 x 2 (e) G1x2 = - x + 3 3 55. Under what circumstances is a linear function f 1x2 = mx + b odd? Can a linear function ever be even? 56. Explain how the graph of f 1x2 = mx + b can be used to solve mx + b 7 0. Retain Your Knowledge Problems 57–60 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 57. Graph x2 - 4x + y2 + 10y - 7 = 0. 58. If f(x) = 2x + B and f(5) = 8, what is the value of B? x - 3 59. Find the average rate of change of f(x) = 3x2 - 5x from 1 to 3. x … 0 x2 60. Graph g(x) = e 2x + 1 x 7 0 ‘Are You Prepared?’ Answers 1. 2. y 22 21 21 1 2 x 2 3 3. 18 4. 5506 5. 0 6. True 284 CHAPTER 4 Linear and Quadratic Functions 4.2 Building Linear Models from Data PREPARING FOR THIS SECTION Before getting started, review the following: r Rectangular Coordinates (Section 2.1, pp. 150–151) r Functions (Section 3.1, pp. 199–208) r Lines (Section 2.3, pp. 167–175) Now Work the ‘Are You Prepared?’ problems on page 287. OBJECTIVES 1 Draw and Interpret Scatter Diagrams (p. 284) 2 Distinguish between Linear and Nonlinear Relations (p. 285) 3 Use a Graphing Utility to Find the Line of Best Fit (p. 286) 1 Draw and Interpret Scatter Diagrams In Section 4.1, we built linear models from verbal descriptions. Linear models can also be constructed by fitting a linear function to data. The first step is to plot the ordered pairs using rectangular coordinates. The resulting graph is a scatter diagram. EX AMPLE 1 Drawing and Interpreting a Scatter Diagram In baseball, the on-base percentage for a team represents the percentage of time that the players safely reach base. The data given in Table 6 represent the number of runs scored y and the on-base percentage x for teams in the National League during the 2013 baseball season. Table 6 Team On-Base Percentage, x Runs Scored, y Arizona 32.3 685 (x, y) (32.3, 685) Atlanta 32.1 688 (32.1, 688) Chicago Cubs 30.0 602 (30.0, 602) Cincinnati 32.7 698 (32.7, 698) Colorado 32.3 706 (32.3, 706) LA Dodgers 32.6 649 (32.6, 649) Miami 29.3 513 (29.3, 513) Milwaukee 31.1 640 (31.1, 640) NY Mets 30.6 619 (30.6, 619) Philadelphia 30.6 610 (30.6, 610) Pittsburgh 31.3 634 (31.3, 634) San Diego 30.8 618 (30.8, 618) San Francisco 32.0 629 (32.0, 629) St. Louis 33.2 783 (33.2, 783) Washington 31.3 656 (31.3, 656) Source: espn.go.com (a) Draw a scatter diagram of the data, treating on-base percentage as the independent variable. (b) Use a graphing utility to draw a scatter diagram. (c) Describe what happens to runs scored as the on-base percentage increases. Solution (a) To draw a scatter diagram, plot the ordered pairs listed in Table 6, with the on-base percentage as the x-coordinate and the runs scored as the y-coordinate. See Figure 6(a). Notice that the points in the scatter diagram are not connected. SECTION 4.2 Building Linear Models from Data 285 (b) Figure 6(b) shows a scatter diagram using a TI-84 Plus C graphing calculator. (c) The scatter diagrams show that as the on-base percentage increases, the number of runs scored also increases. Runs Scored versus On-base Percentage in the National League, 2013 y 850 Runs Scored 800 850 750 700 650 600 550 500 29 450 0 29 29.5 30 30.5 31 31.5 32 On-base Percentage 32.5 33 33.5 x (a) Figure 6 Now Work PROBLEM 34 r (b) 11(a) 2 Distinguish between Linear and Nonlinear Relations Notice that the points in Figure 6 do not follow a perfect linear relation (as they do in Figure 3 in Section 4.1). However, the data do exhibit a linear pattern. There are numerous possible explanations why the data are not perfectly linear, but one easy explanation is the fact that other variables besides on-base percentage (such as number of home runs hit) play a role in determining runs scored. Scatter diagrams are used to help us to see the type of relation that exists between two variables. In this text, we will discuss a variety of different relations that may exist between two variables. For now, we concentrate on distinguishing between linear and nonlinear relations. See Figure 7. Figure 7 (a) Linear y mx b, m 0 EXAMPL E 2 (b) Linear y mx b, m 0 (c) Nonlinear (d) Nonlinear (e) Nonlinear Distinguishing between Linear and Nonlinear Relations Determine whether the relation between the two variables in each scatter diagram in Figure 8 is linear or nonlinear. Figure 8 Solution (a) (a) Linear (b) (b) Nonlinear Now Work PROBLEM (c) (c) Nonlinear 5 (d) (d) Nonlinear r CHAPTER 4 Linear and Quadratic Functions This section considers data whose scatter diagrams suggest that a linear relation exists between the two variables. Suppose that the scatter diagram of a set of data appears to indicate a linear relationship, as in Figure 7(a) or (b). We might want to model the data by finding an equation of a line that relates the two variables. One way to obtain a model for such data is to draw a line through two points on the scatter diagram and determine the equation of the line. EX A MPLE 3 Finding a Model for Linearly Related Data Use the data in Table 6 from Example 1. (a) Select two points and find an equation of the line containing the points. (b) Graph the line on the scatter diagram obtained in Example 1(a). Solution (a) Select two points, say 130.6, 6102 and 132.1, 6882 . The slope of the line joining the points 130.6, 6102 and 132.1, 6882 is m = 78 688 - 610 = = 52 32.1 - 30.6 1.5 The equation of the line that has slope 52 and passes through 130.6, 6102 is found using the point–slope form with m = 52, x1 = 30.6, and y1 = 610. y - y1 = m1x - x1 2 Point–slope form of a line y - 610 = 521x - 30.62 x1 = 30.6, y1 = 610, m = 52 y - 610 = 52x - 1591.2 y = 52x - 981.2 The model (b) Figure 9 shows the scatter diagram with the graph of the line found in part (a). Runs Scored versus On-base Percentage in the National League, 2013 y 850 800 Runs Scored 286 750 (32.1, 688) 700 650 600 (30.6, 610) 550 500 0 29 29.5 30 Figure 9 30.5 31 31.5 32 On-base Percentage 32.5 33 33.5 x r Select two other points and complete the solution. Graph the line on the scatter diagram obtained in Figure 6. Now Work PROBLEMS 11(b) AND (c) 3 Use a Graphing Utility to Find the Line of Best Fit The model obtained in Example 3 depends on the selection of points, which will vary from person to person. So the model that we found might be different from the model you found. Although the model in Example 3 appears to fit the data well, SECTION 4.2 Building Linear Models from Data 287 there may be a model that “fits them better.” Do you think your model fits the data better? Is there a line of best fit? As it turns out, there is a method for finding a model that best fits linearly related data (called the line of best fit).* EXAMPL E 4 Finding a Model for Linearly Related Data Use the data in Table 6 from Example 1. (a) Use a graphing utility to find the line of best fit that models the relation between on-base percentage and runs scored. (b) Graph the line of best fit on the scatter diagram obtained in Example 1(b). (c) Interpret the slope. (d) Use the line of best fit to predict the number of runs a team will score if their on-base percentage is 33.1. Solution (a) Graphing utilities contain built-in programs that find the line of best fit for a collection of points in a scatter diagram. Executing the LINear REGression program provides the results shown in Figure 10. This output shows the equation y = ax + b, where a is the slope of the line and b is the y-intercept. The line of best fit that relates on-base percentage to runs scored may be expressed as the line y = 49.40x - 906.29 The model (b) Figure 11 shows the graph of the line of best fit, along with the scatter diagram. (c) The slope of the line of best fit is 49.40, which means that, for every 1 percent increase in the on-base percentage, runs scored increase by 49.40, on average. (d) Letting x = 33.1 in the equation of the line of best fit, we obtain y = 49.40133.12 - 906.29 ≈ 729 runs. Figure 10 r 850 29 450 Now Work 34 Figure 11 PROBLEMS 11(d) AND (e) Does the line of best fit appear to be a good fit? In other words, does it appear to accurately describe the relation between on-base percentage and runs scored? And just how “good” is this line of best fit? Look again at Figure 10. The last line of output is r = 0.896. This number, called the correlation coefficient, r, - 1 … r … 1, is a measure of the strength of the linear relation that exists between two variables. The closer r is to 1, the more nearly perfect the linear relationship is. If r is close to 0, there is little or no linear relationship between the variables. A negative value of r, r 6 0, indicates that as x increases, y decreases; a positive value of r, r 7 0, indicates that as x increases, y does also. The data given in Table 6, which have a correlation coefficient of 0.896, are indicative of a linear relationship with positive slope. 4.2 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Plot the points (1, 5), (2, 6), (3, 9), (1, 12) in the Cartesian plane. Is the relation 5 11, 52, 12, 62, 13, 92, 11, 122 6 a function? Why? (pp. 150 and 199–208) 2. Find an equation of the line containing the points 11, 42 and 13, 82 . (pp. 167–175) Concepts and Vocabulary 3. A is used to help us to see what type of relation, if any, may exist between two variables. 4. True or False The correlation coefficient is a measure of the strength of a linear relation between two variables and must lie between - 1 and 1, inclusive. * We shall not discuss the underlying mathematics of lines of best fit in this text. 288 CHAPTER 4 Linear and Quadratic Functions Skill Building In Problems 5–10, examine the scatter diagram and determine whether the type of relation is linear or nonlinear. 5. y 6. y 35 30 25 20 15 10 5 14 12 10 8 6 4 2 7. 22 −2 0 2 4 6 8 10 12 14 16 x 0 5 10 1520 2530 3540 x 8. 9. 55 10. 25 5 20 35 0 0 35 12 0 0 0 10 45 *In Problems 11–16: (a) Draw a scatter diagram. (b) Select two points from the scatter diagram, and find the equation of the line containing the points selected. (c) Graph the line found in part (b) on the scatter diagram. (d) Use a graphing utility to find the line of best fit. (e) Use a graphing utility to draw the scatter diagram and graph the line of best fit on it. 11. 14. x 3 4 5 6 7 8 y 4 6 7 x -2 -1 0 1 2 y 7 6 3 2 0 12. 9 10 12 14 16 15. x 3 5 7 9 11 6 9 13. 13 y 0 2 3 11 x - 20 - 17 - 15 - 14 - 10 y 100 120 118 130 140 16. x -2 -1 0 1 2 y -4 0 1 4 5 x - 30 - 27 - 25 - 20 - 14 y 10 12 13 13 18 Applications and Extensions 17. Candy The following data represent the weight (in grams) of various candy bars and the corresponding number of calories. Weight, x Calories, y Hershey’s Milk Chocolate® 44.28 230 Nestle’s Crunch® 44.84 230 Butterfinger® 61.30 270 Baby Ruth® 66.45 Almond Joy® 47.33 Candy Bar (d) Graph the line on the scatter diagram drawn in part (a). (e) Use the linear model to predict the number of calories in a candy bar that weighs 62.3 grams. (f) Interpret the slope of the line found in part (c). 18. Raisins The following data represent the weight (in grams) of a box of raisins and the number of raisins in the box. Weight (grams), w Number of Raisins, N 280 42.3 87 220 42.7 91 93 Twix® (with caramel) 58.00 280 42.8 Snickers® 61.12 280 42.4 87 210 42.6 89 Source: Megan Pocius, student at Joliet Junior College 42.4 90 42.3 82 42.5 86 42.7 86 42.5 86 Heath® 39.52 (a) Draw a scatter diagram of the data, treating weight as the independent variable. (b) What type of relation appears to exist between the weight of a candy bar and the number of calories? (c) Select two points and find a linear model that contains the points. Source: Jennifer Maxwell, student at Joliet Junior College SECTION 4.2 Building Linear Models from Data (a) Draw a scatter diagram of the data, treating weight as the independent variable. (b) What type of relation appears to exist between the weight of a box of raisins and the number of raisins? (c) Select two points and find a linear model that contains the points. (d) Graph the line on the scatter diagram drawn in part (b). (e) Use the linear model to predict the number of raisins in a box that weighs 42.5 grams. (f) Interpret the slope of the line found in part (c). (d) Predict the head circumference of a child who is 26 inches tall. (e) What is the height of a child whose head circumference is 17.4 inches? Height, H (inches) Head Circumference, C (inches) 25.25 16.4 25.75 16.9 25 16.9 27.75 17.6 26.5 17.3 27 17.5 26.75 17.3 26.75 17.5 27.5 17.5 19. Video Games and Grade-Point Average Professor Grant Alexander wanted to find a linear model that relates the number of hours a student plays video games each week, h, to the cumulative grade-point average, G, of the student. He obtained a random sample of 10 full-time students at his college and asked each student to disclose the number of hours spent playing video games and the student’s cumulative grade-point average. Hours of Video Games per Week, h Grade-point Average, G 0 3.49 0 3.05 2 3.24 3 2.82 3 3.19 5 2.78 8 2.31 8 2.54 10 2.03 12 2.51 Source: Denise Slucki, student at Joliet Junior College 21. Flight Time and Ticket Price The following data represent nonstop flight time (in minutes) and one-way ticket price (in dollars) for flying from Chicago to various cities on Southwest Airlines. City 20. Height versus Head Circumference A pediatrician wanted to find a linear model that relates a child’s height, H, to head circumference, C. She randomly selects nine children from her practice, measures their height and head circumference, and obtains the data shown. Let H represent the independent variable and C the dependent variable. (a) Use a graphing utility to draw a scatter diagram. (b) Use a graphing utility to find the line of best fit that models the relation between height and head circumference. Express the model using function notation. (c) Interpret the slope. Price Time (minutes), t (dollars), P Atlanta, GA 110 140 Los Angeles, CA 260 199 80 108 125 130 Nashville, TN (a) Explain why the number of hours spent playing video games is the independent variable and cumulative grade-point average is the dependent variable. (b) Use a graphing utility to draw a scatter diagram. (c) Use a graphing utility to find the line of best fit that models the relation between number of hours of video game playing each week and grade-point average. Express the model using function notation. (d) Interpret the slope. (e) Predict the grade-point average of a student who plays video games for 8 hours each week. (f) How many hours of video game playing do you think a student plays whose grade-point average is 2.40? 289 Oklahoma City, OK Omaha, NE 85 89 Ft. Myers, FL 170 175 Phoenix, AZ 225 191 Houston, TX 155 147 Seattle, WA 265 196 65 93 St. Louis, MO Source: Southwest.com, for midmorning flights in May 2014 (a) Use a graphing utility to draw a scatter diagram. (b) Use a graphing utility to find the line of best fit that models the relation between flight time and airfare. Express the model using function notation. (c) Interpret the slope. (d) Predict the airfare for a flight from Chicago to Kansas City, Missouri, if the flight time is 85 minutes. Round to the nearest dollar. (e) Predict the flight time from Chicago to Baltimore, Maryland if the airfare is $120. Round to the nearest minute. 290 CHAPTER 4 Linear and Quadratic Functions Explaining Concepts: Discussion and Writing 22. Maternal Age versus Down Syndrome A biologist would like to know how the age of the mother affects the incidence of Down syndrome. The data to the right represent the age of the mother and the incidence of Down syndrome per 1000 pregnancies. Draw a scatter diagram treating age of the mother as the independent variable. Would it make sense to find the line of best fit for these data? Why or why not? Age of Mother, x Incidence of Down Syndrome, y 33 2.4 34 3.1 23. Find the line of best fit for the ordered pairs 11, 52 and 13, 82 . What is the correlation coefficient for these data? Why is this result reasonable? 35 4 36 5 37 6.7 24. What does a correlation coefficient of 0 imply? 38 8.3 25. Explain why it does not make sense to interpret the y-intercept in Problem 17. 39 10 40 13.3 41 16.7 42 22.2 43 28.6 44 33.3 45 50 26. Refer to Problem 19. Solve G1h2 = 0. Provide an interpretation of this result. Find G102 . Provide an interpretation of this result. Source: Hook, E.B., Journal of the American Medical Association, 249, 2034-2038, 1983. Retain Your Knowledge Problems 27–30 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 27. Find an equation for the line containing the points (–1, 5) and (3, –3). Express your answer using either the general form or the slope-intercept form of the equation of a line, whichever you prefer. x - 1 28. Find the domain of f(x) = 2 . x - 25 29. For f(x) = 5x - 8 and g(x) = x2 - 3x + 4, find (g - f)(x). 30. Write the function whose graph is the graph of y = x2, but shifted to the left 3 units and shifted down 4 units. ‘Are You Prepared?’ Answers 1. No, because the input, 1, corresponds to two different outputs. y 12 9 2. y = 2x + 2 6 3 1 2 3 x 4.3 Quadratic Functions and Their Properties PREPARING FOR THIS SECTION Before getting started, review the following: r Intercepts (Section 2.2, pp. 159–160) r Graphing Techniques: Transformations (Section 3.5, pp. 247–256) r Completing the Square (Section R.5, p. 56) r Quadratic Equations (Section 1.2, pp. 92–99) Now Work the ‘Are You Prepared?’ problems on page 299. OBJECTIVES 1 2 3 4 5 Graph a Quadratic Function Using Transformations (p. 292) Identify the Vertex and Axis of Symmetry of a Quadratic Function (p. 294) Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts (p. 294) Find a Quadratic Function Given Its Vertex and One Other Point (p. 297) Find the Maximum or Minimum Value of a Quadratic Function (p. 298) SECTION 4.3 Quadratic Functions and Their Properties 291 Quadratic Functions Here are some examples of quadratic functions. F1x2 = 3x2 - 5x + 1 DEFINITION g1x2 = - 6x2 + 1 H1x2 = 1 2 2 x + x 2 3 A quadratic function is a function of the form f1x2 = ax2 + bx + c where a, b, and c are real numbers and a ≠ 0. The domain of a quadratic function is the set of all real numbers. In Words A quadratic function is a function defined by a second-degree polynomial in one variable. Many applications require a knowledge of quadratic functions. For example, suppose that Texas Instruments collects the data shown in Table 7, which relate the number of calculators sold to the price p (in dollars) per calculator. Since the price of a product determines the quantity that will be purchased, we treat price as the independent variable. The relationship between the number x of calculators sold and the price p per calculator is given by the linear equation x = 21,000 - 150p Price p per Calculator, Number of (in dollars) Calculators, x Table 7 60 12,000 65 11,250 70 10,500 75 9,750 80 9,000 85 8,250 90 7,500 Then the revenue R derived from selling x calculators at the price p per calculator is equal to the unit selling price p of the calculator times the number x of units actually sold. That is, R = xp R 1p2 = 121,000 - 150p2p x = 21,000 - 150p = - 150p2 + 21,000p So the revenue R is a quadratic function of the price p. Figure 12 illustrates the graph of this revenue function, whose domain is 0 … p … 140, since both x and p must be nonnegative. R 800,000 Revenue (dollars) 700,000 600,000 500,000 400,000 300,000 200,000 100,000 Figure 12 R(p) = - 150p2 + 21,000p 0 14 28 42 56 70 84 98 112 126 140 Price per calculator (dollars) p 292 CHAPTER 4 Linear and Quadratic Functions A second situation in which a quadratic function appears involves the motion of a projectile. Based on Newton’s Second Law of Motion (force equals mass times acceleration, F = ma), it can be shown that, ignoring air resistance, the path of a projectile propelled upward at an inclination to the horizontal is the graph of a quadratic function. See Figure 13 for an illustration. 1 Graph a Quadratic Function Using Transformations Figure 13 Path of a cannonball We know how to graph the square function f1x2 = x2. Figure 14 shows the graph of 1 three functions of the form f1x2 = ax2, a 7 0, for a = 1, a = , and a = 3. Note 2 that the larger the value of a, the “narrower” the graph is, and the smaller the value of a, the “wider” the graph is. Figure 15 shows the graphs of f1x2 = ax2 for a 6 0. Notice that these graphs are reflections about the x-axis of the graphs in Figure 14. Based on the results of these two figures, general conclusions can be drawn about the graph of f1x2 = ax2. First, as 0 a 0 increases, the graph is stretched vertically (becomes “taller”), and as 0 a 0 gets closer to zero, the graph is compressed vertically (becomes “shorter”). Second, if a is positive, the graph opens “up,” and if a is negative, the graph opens “down.” y y 4 f (x ) = 3x 2 4 f (x ) = x 2 f (x ) = 1–2 x 2 –4 –4 4 4 x x f (x ) = – 1–2 x 2 –4 f (x) = –x 2 f (x) = – 3x 2 –4 Figure 14 Axis of symmetry Vertex is highest point Vertex is lowest point Axis of symmetry D Opens up E Opens down a.0 a,0 Figure 16 Graphs of a quadratic function, f 1x2 = ax2 + bx + c, a ≠ 0 EX AMPLE 1 Figure 15 The graphs in Figures 14 and 15 are typical of the graphs of all quadratic functions, which are called parabolas.* Refer to Figure 16, where two parabolas are pictured. The one on the left opens up and has a lowest point; the one on the right opens down and has a highest point. The lowest or highest point of a parabola is called the vertex. The vertical line passing through the vertex in each parabola in Figure 16 is called the axis of symmetry (usually abbreviated to axis) of the parabola. Because the parabola is symmetric about its axis, the axis of symmetry of a parabola can be used to find additional points on the parabola. The parabolas shown in Figure 16 are the graphs of a quadratic function f1x2 = ax2 + bx + c, a ≠ 0. Notice that the coordinate axes are not included in the figure. Depending on the values of a, b, and c, the axes could be placed anywhere. The important fact is that the shape of the graph of a quadratic function will look like one of the parabolas in Figure 16. In the following example, techniques from Section 3.5 are used to graph a quadratic function f 1x2 = ax2 + bx + c, a ≠ 0. The method of completing the square is used to write the function f in the form f1x2 = a1x - h2 2 + k. Graphing a Quadratic Function Using Transformations Graph the function f1x2 = 2x2 + 8x + 5. Find the vertex and axis of symmetry. * Parabolas will be studied using a geometric definition later in this text. SECTION 4.3 Quadratic Functions and Their Properties Solution 293 Begin by completing the square on the right side. f1x2 = 2x2 + 8x + 5 = 21x2 + 4x2 + 5 Factor out the 2 from 2x 2 + 8x. = 21x2 + 4x + 42 + 5 - 8 Complete the square of x 2 + 4x by adding 4. Notice that the factor of 2 requires that 8 be added and subtracted. = 21x + 22 2 - 3 The graph of f can be obtained from the graph of y = x2 in three stages, as shown in Figure 17. Now compare this graph to the graph in Figure 16(a). The graph of f1x2 = 2x2 + 8x + 5 is a parabola that opens up and has its vertex (lowest point) at 1 - 2, - 32. Its axis of symmetry is the line x = - 2. y 3 y 3 (⫺1, 1) ⫺2 (⫺1, 2) (1, 1) (0, 0) 2 x (a) y ⫽ x 2 (1, 2) ⫺3 (0, 0) ⫺3 Figure 17 y 3 3 (⫺3, 2) x (⫺1, 2) ⫺3 (⫺2, 0) 3 x Replace x by x ⫹ 2; Shift left 2 units (b) y ⫽ 2 x 2 3 (⫺1, ⫺1) (⫺3, ⫺1) ⫺3 ⫺3 Multiply by 2; Vertical stretch Axis of y Symmetry3 x ⫽ ⫺2 (c) y ⫽ 2 (x ⫹ 2) 2 Subtract 3; Shift down 3 units x ⫺3 Vertex (⫺2, ⫺3) (d) y ⫽ 2(x ⫹ 2)2 ⫺ 3 r Now Work PROBLEM 25 The method used in Example 1 can be used to graph any quadratic function f1x2 = ax2 + bx + c, a ≠ 0, as follows: f1x2 = ax2 + bx + c = aax2 + b xb + c a Factor out a from ax 2 + bx. = aax2 + b b2 b2 x + b + c aa b a 4a2 4a2 b2 Complete the square by adding 2 . 4a Look closely at this step! = aax + b 2 b2 b + c 2a 4a Factor ; simplify. = aax + b 2 4ac - b2 b + 2a 4a c - 4a b2 4ac - b 2 b2 = c# = 4a 4a 4a 4a These results lead to the following conclusion: If h = - 4ac - b2 b and k = , then 2a 4a f1x2 = ax2 + bx + c = a1x - h2 2 + k (1) The graph of f1x2 = a1x - h2 2 + k is the parabola y = ax2 shifted horizontally h units (replace x by x - h) and vertically k units (add k). As a result, the vertex is at 1h, k2 , and the graph opens up if a 7 0 and down if a 6 0. The axis of symmetry is the vertical line x = h. 294 CHAPTER 4 Linear and Quadratic Functions For example, compare equation (1) with the solution given in Example 1. f 1x2 = 21x + 22 2 - 3 = 21x - 1 - 22 2 2 + 1 - 32 = a1x - h2 2 + k Because a = 2, the graph opens up. Also, because h = - 2 and k = - 3, its vertex is at 1 - 2, - 32. 2 Identify the Vertex and Axis of Symmetry of a Quadratic Function We do not need to complete the square to obtain the vertex. In almost every case, it is easier to obtain the vertex of a quadratic function f by remembering that b its x-coordinate is h = - . The y-coordinate k can then be found by evaluating f 2a b b at - . That is, k = f a - b . 2a 2a Properties of the Graph of a Quadratic Function f1x2 = ax2 + bx + c Vertex = a - b b , f a- b b 2a 2a a ≠ 0 Axis of symmetry: the vertical line x = - b (2) 2a Parabola opens up if a 7 0; the vertex is a minimum point. Parabola opens down if a 6 0; the vertex is a maximum point. EX AMPLE 2 Locating the Vertex without Graphing Without graphing, locate the vertex and axis of symmetry of the parabola defined by f1x2 = - 3x2 + 6x + 1. Does it open up or down? Solution For this quadratic function, a = - 3, b = 6, and c = 1. The x-coordinate of the vertex is h = - b 6 = = 1 2a 21 - 32 The y-coordinate of the vertex is k = f a- b b = f112 = - 3112 2 + 6112 + 1 = 4 2a The vertex is located at the point 11, 42 . The axis of symmetry is the line x = 1. Because a = - 3 6 0, the parabola opens down. r 3 Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts The location of the vertex and intercepts, along with knowledge of whether the graph opens up or down, usually provides enough information to graph f1x2 = ax2 + bx + c, a ≠ 0. The y-intercept is the value of f at x = 0; that is, the y-intercept is f102 = c. The x-intercepts, if there are any, are found by solving the quadratic equation ax2 + bx + c = 0 SECTION 4.3 Quadratic Functions and Their Properties 295 This equation has two, one, or no real solutions, depending on whether the discriminant b2 - 4ac is positive, 0, or negative. Depending on the value of the discriminant, the graph of f has x-intercepts, as follows: The x-Intercepts of a Quadratic Function 1. If the discriminant b2 - 4ac 7 0, the graph of f1x2 = ax2 + bx + c has two distinct x-intercepts so it crosses the x-axis in two places. 2. If the discriminant b2 - 4ac = 0, the graph of f 1x2 = ax2 + bx + c has one x-intercept so it touches the x-axis at its vertex. 3. If the discriminant b2 - 4ac 6 0, the graph of f1x2 = ax2 + bx + c has no x-intercepts so it does not cross or touch the x-axis. Figure 18 illustrates these possibilities for parabolas that open up. Axis of symmetry x=– b 2a y Axis of symmetry x=– b 2a y x-intercept x -intercept – b ,f – b 2a 2a ( Figure 18 f(x) = ax + bx + c, a 7 0 ( D 2 EXAMPL E 3 b2 – x Axis of symmetry x=– b 2a y x x -intercept x b b ) (– 2a , f (– 2a )) – b ,0 2a ( )) 4ac > 0 E b 2 – 4ac = 0 One x-intercept Two x-intercepts F b 2 – 4ac < 0 No x-intercepts Graphing a Quadratic Function Using Its Vertex, Axis, and Intercepts (a) Use the information from Example 2 and the locations of the intercepts to graph f1x2 = - 3x2 + 6x + 1. (b) Determine the domain and the range of f. (c) Determine where f is increasing and where it is decreasing. Solution (a) In Example 2, we found the vertex to be at 11, 42 and the axis of symmetry to be x = 1. The y-intercept is found by letting x = 0. The y-intercept is f102 = 1. The x-intercepts are found by solving the equation f1x2 = 0. This results in the equation - 3x2 + 6x + 1 = 0 a = - 3, b = 6, c = 1 The discriminant b - 4ac = 162 - 41 - 32 112 = 36 + 12 = 48 7 0, so the equation has two real solutions and the graph has two x-intercepts. Use the quadratic formula to find that 2 Axis of symmetry x 1 y (1, 4) 4 (0, 1) 4 (0.15, 0) (2, 1) 4 x (2.15, 0) Figure 19 f(x) = - 3x2 + 6x + 1 x = 2 - b + 2b2 - 4ac - 6 + 248 - 6 + 423 = = ≈ - 0.15 2a -6 -6 and - 6 - 248 - 6 - 423 - b - 2b2 - 4ac = = ≈ 2.15 2a -6 -6 The x-intercepts are approximately - 0.15 and 2.15. The graph is illustrated in Figure 19. Notice how we used the y-intercept and the axis of symmetry, x = 1, to obtain the additional point 12, 12 on the graph. x = 296 CHAPTER 4 Linear and Quadratic Functions (b) The domain of f is the set of all real numbers. Based on the graph, the range of f is the interval 1 - q , 44. (c) The function f is increasing on the interval 1 - q , 12 and decreasing on the interval 11, q 2. r Graph the function in Example 3 by completing the square and using transformations. Which method do you prefer? Now Work PROBLEM 33 If the graph of a quadratic function has only one x-intercept or no x-intercepts, it is usually necessary to plot an additional point to obtain the graph. EX AMPLE 4 Graphing a Quadratic Function Using Its Vertex, Axis, and Intercepts (a) Graph f1x2 = x2 - 6x + 9 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of f. (c) Determine where f is increasing and where it is decreasing. Solution (a) For f1x2 = x2 - 6x + 9, note that a = 1, b = - 6, and c = 9. Because a = 1 7 0, the parabola opens up. The x-coordinate of the vertex is h = - Axis of symmetry y x3 b -6 = = 3 2a 2112 The y-coordinate of the vertex is (0, 9) k = f 132 = 132 2 - 6132 + 9 = 0 (6, 9) 6 3 (3, 0) 6 x Figure 20 f (x) = x2 - 6x + 9 The vertex is at 13, 02 . The axis of symmetry is the line x = 3. The y-intercept is f102 = 9. Since the vertex 13, 02 lies on the x-axis, the graph touches the x-axis at the x-intercept. By using the axis of symmetry and the y-intercept at (0, 9), we can locate the additional point 16, 92 on the graph. See Figure 20. (b) The domain of f is the set of all real numbers. Based on the graph, the range of f is the interval 3 0, q 2. (c) The function f is decreasing on the interval 1 - q , 32 and increasing on the interval 13, q 2. r Now Work EX AMPLE 5 PROBLEM 39 Graphing a Quadratic Function Using Its Vertex, Axis, and Intercepts (a) Graph f1x2 = 2x2 + x + 1 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of f. (c) Determine where f is increasing and where it is decreasing. Solution (a) For f1x2 = 2x2 + x + 1, we have a = 2, b = 1, and c = 1. Because a = 2 7 0, the parabola opens up. The x-coordinate of the vertex is h = - b 1 = 2a 4 SECTION 4.3 Quadratic Functions and Their Properties NOTE In Example 5, since the vertex is above the x-axis and the parabola opens up, we can conclude that the graph of the quadratic function will have no x-intercepts. ■ 2 1 ( – 1–2 , 1) The y-coordinate of the vertex is 1 1 1 7 k = f a - b = 2a b + a - b + 1 = 4 16 4 8 1 7 1 The vertex is at a - , b . The axis of symmetry is the line x = - . The 4 8 4 y-intercept is f 102 = 1. The x-intercept(s), if any, satisfy the equation 2x2 + x + 1 = 0. The discriminant b2 - 4ac = 112 2 - 4122 112 = - 7 6 0. This equation has no real solutions, which means the graph has no x-intercepts. 1 Use the point (0, 1) and the axis of symmetry x = - to locate the additional 4 1 point a - , 1b on the graph. See Figure 21. 2 y Axis of symmetry x – 1–4 297 (b) The domain of f is the set of all real numbers. Based on the graph, the range of f 7 is the interval c , q b . 8 1 (c) The function f is decreasing on the interval a - q , - b and is increasing on the 4 1 interval a - , q b . 4 (0, 1) ( – 1–4 , 7–8 ) –1 x 1 r Figure 21 f (x) = 2x + x + 1 2 Now Work PROBLEM 43 4 Find a Quadratic Function Given Its Vertex and One Other Point If the vertex 1h, k2 and one additional point on the graph of a quadratic function f 1x2 = ax2 + bx + c, a ≠ 0, are known, then f 1x2 = a1x - h2 2 + k (3) can be used to obtain the quadratic function. EXAMPL E 6 Finding the Quadratic Function Given Its Vertex and One Other Point Determine the quadratic function whose vertex is 11, - 52 and whose y-intercept is - 3. Solution The vertex is 11, - 52, so h = 1 and k = - 5. Substitute these values into equation (3). y h = 1, k = - 5 f1x2 = a1x - 12 2 - 5 4 (0, –3) f1x2 = a1x - 12 - 5 To determine the value of a, use the fact that f102 = - 3 (the y-intercept). 8 –1 Equation (3) 2 12 –2 f1x2 = a1x - h2 2 + k - 3 = a10 - 12 2 - 5 1 2 3 x = 0, y = f1 02 = - 3 -3 = a - 5 4 x a = 2 –4 (1, –5) –8 The quadratic function we seek is f1x2 = a1x - h2 2 + k = 21x - 12 2 - 5 = 2x2 - 4x - 3 Figure 22 f 1x2 = 2x2 - 4x - 3 See Figure 22. Now Work PROBLEM 49 r 298 CHAPTER 4 Linear and Quadratic Functions 5 Find the Maximum or Minimum Value of a Quadratic Function The graph of a quadratic function f1x2 = ax2 + bx + c a ≠ 0 b b , f a - b b . This vertex is the highest point on the 2a 2a graph if a 6 0 and the lowest point on the graph if a 7 0. If the vertex is the highest b point 1a 6 02, then f a - b is the maximum value of f. If the vertex is the 2a b lowest point 1a 7 02, then f a - b is the minimum value of f. 2a is a parabola with vertex at a - EX AMPLE 7 Finding the Maximum or Minimum Value of a Quadratic Function Determine whether the quadratic function f1x2 = x2 - 4x - 5 has a maximum or a minimum value. Then find the maximum or minimum value. Solution Compare f 1x2 = x2 - 4x - 5 to f 1x2 = ax2 + bx + c. Then a = 1, b = - 4, and c = - 5. Because a 7 0, the graph of f opens up, which means the vertex is a minimum point. The minimum value occurs at x = - b -4 4 = = = 2 2a 2112 2 c a = 1, b = - 4 The minimum value is f a- b b = f122 = 22 - 4122 - 5 = 4 - 8 - 5 = - 9 2a Now Work PROBLEM 57 SUMMARY Steps for Graphing a Quadratic Function f(x) = ax2 + bx + c, a ≠ 0 Option 1 STEP 1: Complete the square in x to write the quadratic function in the form f1x2 = a1x - h2 2 + k. STEP 2: Graph the function in stages using transformations. Option 2 STEP 1: Determine whether the parabola opens up 1a 7 02 or down 1a 6 02. b b STEP 2: Determine the vertex a - , f a - b b . 2a 2a b STEP 3: Determine the axis of symmetry, x = - . 2a STEP 4: Determine the y-intercept, f 102 , and the x-intercepts, if any. (a) If b2 - 4ac 7 0, the graph of the quadratic function has two x-intercepts, which are found by solving the equation ax2 + bx + c = 0. (b) If b2 - 4ac = 0, the vertex is the x-intercept. (c) If b2 - 4ac 6 0, there are no x-intercepts. STEP 5: Determine an additional point by using the y-intercept and the axis of symmetry. STEP 6: Plot the points and draw the graph. r 299 SECTION 4.3 Quadratic Functions and Their Properties 4.3 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. List the intercepts of the equation y = x2 - 9. (pp. 159–160) 3. To complete the square of x2 - 5x, you add the number . (p. 56) 2. Find the real solutions of the equation 2x + 7x - 4 = 0. (pp. 92–99) 2 4. To graph y = 1x - 42 2, you shift the graph of y = x2 to the a distance of units. (pp. 247–256) Concepts and Vocabulary 5. The graph of a quadratic function is called a(n) 11. If b2 - 4ac 7 0, which of the following conclusions can be made about the graph of f(x) = ax2 + bx + c, a ≠ 0? (a) The graph has two distinct x-intercepts. (b) The graph has no x-intercepts. (c) The graph has three distinct x-intercepts. (d) The graph has one x-intercept. . 6. The vertical line passing through the vertex of a parabola is called the . 7. The x-coordinate of the vertex of f 1x2 = ax2 + bx + c, a ≠ 0, is . 8. True or False The graph of f 1x2 = 2x2 + 3x - 4 opens up. 12. If the graph of f(x) = ax2 + bx + c, a ≠ 0, has a maximum value at its vertex, which of the following conditions must be true? b b (a) 7 0 (b) 6 0 2a 2a (d) a 6 0 (c) a 7 0 9. True or False The y-coordinate of the vertex of f 1x2 = - x2 + 4x + 5 is f 122 . 10. True or False If the discriminant b2 - 4ac = 0, the graph of f 1x2 = ax2 + bx + c, a ≠ 0, will touch the x-axis at its vertex. Skill Building In Problems 13–20, match each graph to one the following functions. 13. f 1x2 = x2 - 1 17. f 1x2 = x2 - 2x + 2 A. 14. f 1x2 = - x2 - 1 18. f 1x2 = x2 + 2x B. y 15. f 1x2 = x2 - 2x + 1 19. f 1x2 = x2 - 2x C. y 3 20. f 1x2 = x2 + 2x + 2 D. y 2 2 16. f 1x2 = x2 + 2x + 1 y 3 (1, 1) 2 2 x 2 (1, 0) E. 2 x 1 2 2 x (0, 1) 3 2 (1, 1) 2 F. y 1 2 x G. y 1 (1, 0) 3 x 1 2 x 1 y 2 2 (1, 1) 1 2 H. y 3 3 x 2 (0, 1) x 2 (1, 1) In Problems 21–32, graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflecting). [Hint: If necessary, write f in the form f 1x2 = a1x - h2 2 + k.] 1 22. f 1x2 = 2x2 + 4 23. f 1x2 = 1x + 22 2 - 2 24. f 1x2 = 1x - 32 2 - 10 21. f 1x2 = x2 4 25. f 1x2 = x2 + 4x + 2 26. f 1x2 = x2 - 6x - 1 27. f 1x2 = 2x2 - 4x + 1 28. f 1x2 = 3x2 + 6x 1 2 4 29. f 1x2 = - x2 - 2x 30. f 1x2 = - 2x2 + 6x + 2 31. f 1x2 = x2 + x - 1 32. f 1x2 = x2 + x - 1 2 3 3 In Problems 33–48, (a) graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of the function. (c) Determine where the function is increasing and where it is decreasing. 33. f 1x2 = x2 + 2x 37. f 1x2 = x + 2x - 8 2 34. f 1x2 = x2 - 4x 38. f 1x2 = x - 2x - 3 2 35. f 1x2 = - x2 - 6x 39. f 1x2 = x + 2x + 1 2 36. f 1x2 = - x2 + 4x 40. f 1x2 = x2 + 6x + 9 300 CHAPTER 4 Linear and Quadratic Functions 41. f 1x2 = 2x2 - x + 2 45. f 1x2 = 3x + 6x + 2 2 42. f 1x2 = 4x2 - 2x + 1 43. f 1x2 = - 2x2 + 2x - 3 46. f 1x2 = 2x + 5x + 3 47. f 1x2 = - 4x - 6x + 2 2 2 In Problems 49–54, determine the quadratic function whose graph is given. 49. 50. y y 2 –2 y 6 Vertex: (–3, 5) 4 4 2 –6 Vertex: (2, 1) –3 1 2 (0, –1) Vertex: (–1, –2) 51. (0, 5) 1 x –1 48. f 1x2 = 3x2 - 8x + 2 8 1 –3 44. f 1x2 = - 3x2 + 3x - 2 –2 x (0, –4) –1 1 2 3 4 5 x –8 52. 53. y 4 Vertex: (2, 3) 6 2 –1 1 (0, –1) 3 y 54. y 8 Vertex: (–2, 6) (3, 5) 4 4 5 x 6 2 2 –3 –4 x 3 –1 (–4, –2) –1 1 x –2 –4 –2 –4 Vertex: (1, –3) In Problems 55–62, determine, without graphing, whether the given quadratic function has a maximum value or a minimum value, and then find the value. 55. f 1x2 = 2x2 + 12x 59. f 1x2 = - x + 10x - 4 2 56. f1x2 = - 2x2 + 12x 60. f 1x2 = - 2x + 8x + 3 2 57. f 1x2 = 2x2 + 12x - 3 61. f 1x2 = - 3x + 12x + 1 2 58. f 1x2 = 4x2 - 8x + 3 62. f 1x2 = 4x2 - 4x Mixed Practice In Problems 63–70, (a) graph each function. (b) Determine the domain and the range of the function. (c) Determine where the function is increasing and where it is decreasing. 2 3 63. f(x) = x2 - 2x - 15 64. g(x) = x2 - 2x - 8 65. h(x) = - x + 4 66. f (x) = x - 2 5 2 67. g(x) = - 2(x - 3)2 + 2 68. h(x) = - 3(x + 1)2 + 4 69. f(x) = 2x2 + x + 1 70. F(x) = - 4x2 + 20x - 25 Applications and Extensions 71. The graph of the function f 1x2 = ax2 + bx + c has vertex at 10, 22 and passes through the point 11, 82. Find a, b, and c. 72. The graph of the function f 1x2 = ax2 + bx + c has vertex at 11, 42 and passes through the point 1 - 1, - 82. Find a, b, and c. In Problems 73–78, for the given functions f and g: (a) Graph f and g on the same Cartesian plane. (b) Solve f 1x2 = g1x2. (c) Use the result of part (b) to label the points of intersection of the graphs of f and g. (d) Shade the region for which f 1x2 7 g1x2 —that is, the region below f and above g. 73. f 1x2 = 2x - 1; g1x2 = x2 - 4 75. f 1x2 = - x + 4; g1x2 = - 2x + 1 2 77. f 1x2 = - x2 + 5x; g1x2 = x2 + 3x - 4 74. f 1x2 = - 2x - 1; g1x2 = x2 - 9 76. f 1x2 = - x2 + 9; g1x2 = 2x + 1 78. f 1x2 = - x2 + 7x - 6; g1x2 = x2 + x - 6 Answer Problems 79 and 80 using the following: A quadratic function of the form f 1x2 = ax2 + bx + c with b2 - 4ac 7 0 may also be written in the form f 1x2 = a1x - r1 2 1x - r2 2, where r1 and r2 are the x-intercepts of the graph of the quadratic function. 79. (a) Find a quadratic function whose x-intercepts are - 3 and 1 with a = 1; a = 2; a = - 2; a = 5. (b) How does the value of a affect the intercepts? (c) How does the value of a affect the axis of symmetry? (d) How does the value of a affect the vertex? (e) Compare the x-coordinate of the vertex with the midpoint of the x-intercepts. What might you conclude? 80. (a) Find a quadratic function whose x-intercepts are - 5 and 3 with a = 1; a = 2; a = - 2; a = 5. (b) How does the value of a affect the intercepts? (c) How does the value of a affect the axis of symmetry? (d) How does the value of a affect the vertex? (e) Compare the x-coordinate of the vertex with the midpoint of the x-intercepts. What might you conclude? SECTION 4.3 Quadratic Functions and Their Properties 81. Suppose that f 1x2 = x2 + 4x - 21. (a) What is the vertex of f? (b) What are the x-intercepts of the graph of f ? (c) Solve f 1x2 = - 21 for x. What points are on the graph of f ? (d) Use the information obtained in parts (a)–(c) to graph f 1x2 = x2 + 4x - 21. 82. Suppose that f 1x2 = x2 + 2x - 8. (a) What is the vertex of f ? (b) What are the x-intercepts of the graph of f ? (c) Solve f 1x2 = - 8 for x. What points are on the graph of f ? (d) Use the information obtained in parts (a)–(c) to graph f 1x2 = x2 + 2x - 8. 83. Find the point on the line y = x that is closest to the point 13, 12. [Hint: Express the distance d from the point to the line as a function of x, and then find the minimum value of 3d1x2 4 2. 84. Find the point on the line y = x + 1 that is closest to the point 14, 12. 85. Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that when the unit price is p dollars, the revenue R (in dollars) is R 1p2 = - 4p2 + 4000p What unit price should be established for the dryer to maximize revenue? What is the maximum revenue? 86. Maximizing Revenue The John Deere company has found that the revenue, in dollars, from sales of riding mowers is a function of the unit price p, in dollars, that it charges. If the revenue R is 1 R 1p2 = - p2 + 1900p 2 what unit price p should be charged to maximize revenue? What is the maximum revenue? 87. Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the marginal cost C (in dollars) to produce x thousand digital music players is given by the function C 1x2 = x2 - 140x + 7400 (a) How many players should be produced to minimize the marginal cost? (b) What is the minimum marginal cost? 88. Minimizing Marginal Cost (See Problem 87.) The marginal cost C (in dollars) of manufacturing x cell phones (in thousands) is given by C 1x2 = 5x2 - 200x + 4000 301 (a) How many cell phones should be manufactured to minimize the marginal cost? (b) What is the minimum marginal cost? 89. Business The monthly revenue R achieved by selling x wristwatches is figured to be R 1x2 = 75x - 0.2x2. The monthly cost C of selling x wristwatches is C 1x2 = 32x+ 1750 (a) How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as P 1x2 = R 1x2 - C 1x2. What is the profit function? (c) How many wristwatches must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a quadratic function is a reasonable model for revenue. 90. Business The daily revenue R achieved by selling x boxes of candy is figured to be R 1x2 = 9.5x - 0.04x2. The daily cost C of selling x boxes of candy is C 1x2 = 1.25x + 250. (a) How many boxes of candy must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as P 1x2 = R 1x2 - C 1x2. What is the profit function? (c) How many boxes of candy must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a quadratic function is a reasonable model for revenue. 91. Stopping Distance An accepted relationship between stopping distance, d (in feet), and the speed of a car, v (in mph), is d = 1.1v + 0.06v2 on dry, level concrete. (a) How many feet will it take a car traveling 45 mph to stop on dry, level concrete? (b) If an accident occurs 200 feet ahead of you, what is the maximum speed you can be traveling to avoid being involved? (c) What might the term 1.1v represent? 92. Birth Rate of Unmarried Women In the United States, the birth rate B of unmarried women (births per 1000 unmarried women) for women whose age is a is modeled by the function B(a) = - 0.30a2 + 16.26a - 158.90 (a) What is the age of unmarried women with the highest birth rate? (b) What is the highest birth rate of unmarried women? (c) Evaluate and interpret B 1402. Source: National Vital Statistics System, 2013 93. Let f 1x2 = ax2 + bx + c, where a, b, and c are odd integers. If x is an integer, show that f 1x2 must be an odd integer. [Hint: x is either an even integer or an odd integer.] Explaining Concepts: Discussion and Writing 94. Make up a quadratic function that opens down and has only one x-intercept. Compare yours with others in the class. What are the similarities? What are the differences? 97. State the circumstances that cause the graph of a quadratic function f 1x2 = ax2 + bx + c to have no x-intercepts. 95. On one set of coordinate axes, graph the family of parabolas f 1x2 = x2 + 2x + c for c = - 3, c = 0, and c = 1. Describe the characteristics of a member of this family. 98. Why does the graph of a quadratic function open up if a 7 0 and down if a 6 0? 99. Can a quadratic function have a range of 1 - q , q 2? Justify your answer. 96. On one set of coordinate axes, graph the family of parabolas f 1x2 = x2 + bx + 1 for b = - 4, b = 0, and b = 4. Describe the general characteristics of this family. 100. What are the possibilities for the number of times the graphs of two different quadratic functions intersect? 302 CHAPTER 4 Linear and Quadratic Functions Retain Your Knowledge Problems 101–104 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 101. Determine whether x2 + 4y2 = 16 is symmetric respect to the x-axis, the y-axis, and/or the origin. 102. Solve the inequality 27 - x Ú 5x + 3. Write the solution in both set notation and interval notation. 103. Find the center and radius of the circle x2 + y2 - 10x + 4y + 20 = 0. 104. Write the function whose graph is the graph of y = 2x, but reflected about the y-axis. ‘Are You Prepared?’ Answers 1. 10, - 92, 1 - 3, 02, 13, 02 1 2. e - 4, f 2 3. 25 4 4. right; 4 4.4 Build Quadratic Models from Verbal Descriptions and from Data PREPARING FOR THIS SECTION Before getting started, review the following: r Problem Solving (Section 1.7, pp. 134–140) r Building Linear Models from Data (Section 4.2, pp. 284–287) Now Work the ‘Are You Prepared?’ problems on page 307. OBJECTIVES 1 Build Quadratic Models from Verbal Descriptions (p. 302) 2 Build Quadratic Models from Data (p. 306) In this section we will first discuss models in the form of a quadratic function when a verbal description of the problem is given. We end the section by fitting a quadratic function to data, which is another form of modeling. When a mathematical model is in the form of a quadratic function, the properties of the graph of the function can provide important information about the model. In particular, we can use the quadratic function to determine the maximum or minimum value of the function. The fact that the graph of a quadratic function has a maximum or minimum value enables us to answer questions involving optimization—that is, finding the maximum or minimum values in models. 1 Build Quadratic Models from Verbal Descriptions In economics, revenue R, in dollars, is defined as the amount of money received from the sale of an item and is equal to the unit selling price p, in dollars, of the item times the number x of units actually sold. That is, R = xp The Law of Demand states that p and x are related: As one increases, the other decreases. The equation that relates p and x is called the demand equation. When the demand equation is linear, the revenue model is a quadratic function. EX AMPLE 1 Maximizing Revenue The marketing department at Texas Instruments has found that when certain calculators are sold at a price of p dollars per unit, the number x of calculators sold is given by the demand equation x = 21,000 - 150p (a) Find a model that expresses the revenue R as a function of the price p. (b) What is the domain of R? (c) What unit price should be used to maximize revenue? (d) If this price is charged, what is the maximum revenue? SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data 303 (e) How many units are sold at this price? (f) Graph R. (g) What price should Texas Instruments charge to collect at least $675,000 in revenue? Solution (a) The revenue is R = xp, where x = 21,000 - 150p. R = xp = 121,000 - 150p2p = - 150p2 + 21,000p The model (b) Because x represents the number of calculators sold, we have x Ú 0, so 21,000 - 150p Ú 0. Solving this linear inequality gives p … 140. In addition, Texas Instruments will charge only a positive price for the calculator, so p 7 0. Combining these inequalities gives the domain of R, which is { p | 0 6 p … 140}. (c) The function R is a quadratic function with a = - 150, b = 21,000, and c = 0. Because a 6 0, the vertex is the highest point on the parabola. The revenue R is a maximum when the price p is p = - 21,000 b = = +70.00 2a c 21 - 1502 a = - 150, b = 21,000 (d) The maximum revenue R is R 1702 = - 1501702 2 + 21,0001702 = +735,000 (e) The number of calculators sold is given by the demand equation x = 21,000 - 150p. At a price of p = +70, x = 21,000 - 1501702 = 10,500 Revenue (dollars) calculators are sold. (f) To graph R, plot the intercept 1140, 02 and the vertex 170, 735 0002 . See Figure 23 for the graph. R 800,000 700,000 600,000 500,000 400,000 300,000 200,000 100,000 (70, 735 000) 0 Figure 23 14 28 42 56 70 84 98 112 126 140 Price per calculator (dollars) p (g) Graph R = 675,000 and R 1p2 = - 150p2 + 21,000p on the same Cartesian plane. See Figure 24. We find where the graphs intersect by solving Revenue (dollars) 675,000 2 150p - 21,000p + 675,000 p2 - 140p + 4500 1p - 502 1p - 902 p = 50 or p R 800,000 700,000 600,000 500,000 400,000 300,000 200,000 100,000 - 150p2 + 21,000p 0 Add 150p 2 - 21,000p to both sides. Divide both sides by 150. 0 Factor. 0 90 Use the Zero-Product Property. (70, 735 000) (90, 675 000) (50, 675 000) 0 Figure 24 = = = = = 14 28 42 56 70 84 98 112 126 140 Price per calculator (dollars) p 304 CHAPTER 4 Linear and Quadratic Functions The graphs intersect at 150, 675 0002 and 190, 675 0002 . Based on the graph in Figure 24, Texas Instruments should charge between $50 and $90 to earn at least $675,000 in revenue. Now Work EX AMPLE 2 PROBLEM r 3 Maximizing the Area Enclosed by a Fence A farmer has 2000 yards of fence to enclose a rectangular field. What are the dimensions of the rectangle that encloses the most area? Solution Figure 25 illustrates the situation. The available fence represents the perimeter of the rectangle. If x is the length and w is the width, then 2x + 2w = 2000 (1) The area A of the rectangle is A = xw To express A in terms of a single variable, solve equation (1) for w and substitute the result in A = xw. Then A involves only the variable x. [You could also solve equation (1) for x and express A in terms of w alone. Try it!] x w w x Figure 25 2x + 2w = 2000 2w = 2000 - 2x 2000 - 2x w = = 1000 - x 2 Then the area A is A = xw = x11000 - x2 = - x2 + 1000x Now, A is a quadratic function of x. A 1x2 = - x2 + 1000x A a = - 1, b = 1000, c = 0 Figure 26 shows the graph of A 1x2 = - x2 + 1000x. Because a 6 0, the vertex is a maximum point on the graph of A. The maximum value occurs at (500, 250 000) x = - 250,000 b 1000 = = 500 2a 21 - 12 The maximum value of A is (0, 0) (1000, 0) 500 1000 x Figure 26 A(x) = - x2 + 1000x Aa - b b = A 15002 = - 5002 + 100015002 = - 250,000 + 500,000 = 250,000 2a The largest rectangle that can be enclosed by 2000 yards of fence has an area of 250,000 square yards. Its dimensions are 500 yards by 500 yards. Now Work EX AMPLE 3 PROBLEM r 7 Analyzing the Motion of a Projectile A projectile is fired from a cliff 500 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 400 feet per second. From physics, the height h of the projectile above the water can be modeled by h 1x2 = - 32x2 + x + 500 14002 2 where x is the horizontal distance of the projectile from the base of the cliff. See Figure 27. h (x) 2500 2000 1500 1000 500 Figure 27 458 1000 2000 3000 4000 5000 x SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data 305 (a) Find the maximum height of the projectile. (b) How far from the base of the cliff will the projectile strike the water? Solution (a) The height of the projectile is given by a quadratic function. h 1x2 = - 32x2 -1 2 + x + 500 = x + x + 500 2 5000 14002 We are looking for the maximum value of h. Because a 6 0, the maximum value occurs at the vertex, whose x-coordinate is x = - b = 2a 1 5000 = = 2500 -1 2 2a b 5000 The maximum height of the projectile is h 125002 = -1 125002 2 + 2500 + 500 = - 1250 + 2500 + 500 = 1750 ft 5000 (b) The projectile will strike the water when the height is zero. To find the distance x traveled, solve the equation h 1x2 = Seeing the Concept -1 2 x + x + 500 = 0 5000 The discriminant of this quadratic equation is Graph -1 2 h(x) = x + x + 500 5000 0 … x … 5500 Use MAXIMUM to find the maximum height of the projectile, and use ROOT or ZERO to find the distance from the base of the cliff to where it strikes the water. Compare your results with those obtained in Example 3. b2 - 4ac = 12 - 4a Then x = - b { 2b2 - 4ac - 1 { 21.4 - 458 = ≈ e 2a 5458 -1 2a b 5000 Discard the negative solution. The projectile will strike the water at a distance of about 5458 feet from the base of the cliff. r Now Work EXAMPL E 4 -1 b 15002 = 1.4 5000 PROBLEM 11 The Golden Gate Bridge The Golden Gate Bridge, a suspension bridge, spans the entrance to San Francisco Bay. Its 746-foot-tall towers are 4200 feet apart. The bridge is suspended from two huge cables more than 3 feet in diameter; the 90-foot-wide roadway is 220 feet above the water. The cables are parabolic in shape* and touch the road surface at the center of the bridge. Find the height of the cable above the road at a distance of 1000 feet from the center. Solution See Figure 28 on page 306. Begin by choosing the placement of the coordinate axes so that the x-axis coincides with the road surface and the origin coincides with the center of the bridge. As a result, the 746-foot towers will be vertical (height 746 - 220 = 526 feet above the road) and located 2100 feet from the center. Also, the cable, which has the shape of a parabola, will extend from the towers, open up, and have its vertex at 10, 02. This choice of placement of the axes enables the equation of the parabola to have the form y = ax2, a 7 0. Note that the points 1 - 2100, 5262 and 12100, 5262 are on the graph. *A cable suspended from two towers is in the shape of a catenary, but when a horizontal roadway is suspended from the cable, the cable takes the shape of a parabola. CHAPTER 4 Linear and Quadratic Functions (2100, 526) (2100, 526) y 526' 306 (0, 0) 2100' 746' x 1000' 220' Figure 28 ? 2100' Use these facts to find the value of a in y = ax2. y = ax2 526 = a121002 2 a = x = 2100, y = 526 526 121002 2 The equation of the parabola is y = 526 x2 121002 2 When x = 1000, the height of the cable is y = 526 110002 2 ≈ 119.3 feet 121002 2 The cable is 119.3 feet above the road at a distance of 1000 feet from the center of the bridge. r Now Work PROBLEM 13 2 Build Quadratic Models from Data In Section 4.2, we found the line of best fit for data that appeared to be linearly related. It was noted that data may also follow a nonlinear relation. Figures 29(a) and (b) show scatter diagrams of data that follow a quadratic relation. y ax 2 bx c, a 0 Figure 29 E X AMPLE 5 (a) y ax 2 bx c, a 0 (b) Fitting a Quadratic Function to Data The data in Table 8 on page 307 represent the percentage D of the population that is divorced for various ages x in 2012. (a) Draw a scatter diagram of the data, treating age as the independent variable. Comment on the type of relation that may exist between age and percentage of the population divorced. (b) Use a graphing utility to find the quadratic function of best fit that models the relation between age and percentage of the population divorced. (c) Use the model found in part (b) to approximate the age at which the percentage of the population divorced is greatest. (d) Use the model found in part (b) to approximate the highest percentage of the population that is divorced. (e) Use a graphing utility to draw the quadratic function of best fit on the scatter diagram. 307 SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data Table 8 Age, x Percentage Divorced, D 22 0.9 27 3.6 32 7.4 37 10.4 42 12.7 50 15.7 60 16.2 70 13.1 80 6.5 Source: United States Statistical Abstract, 2012 Solution 19 15 0 (a) Figure 30 shows the scatter diagram, from which it appears the data follow a quadratic relation, with a 6 0. (b) Execute the QUADratic REGression program to obtain the results shown in Figure 31. The output shows the equation y = ax2 + bx + c. The quadratic function of best fit that models the relation between age and percentage divorced is D1x2 = - 0.0143x2 + 1.5861x - 28.1886 85 The model where x represents age and D represents the percentage divorced. (c) Based on the quadratic function of best fit, the age with the greatest percentage divorced is Figure 30 - b 1.5861 = ≈ 55 years 2a 21 - 0.01432 (d) Evaluate the function D1x2 at x = 55. D1552 = - 0.01431552 2 + 1.58611552 - 28.1886 ≈ 15.8 percent According to the model, 55-year-olds have the highest percentage divorced at 15.8 percent. (e) Figure 32 shows the graph of the quadratic function found in part (b) drawn on the scatter diagram. Figure 31 r 19 15 0 85 Look again at Figure 31. Notice that the output given by the graphing calculator does not include r, the correlation coefficient. Recall that the correlation coefficient is a measure of the strength of a linear relation that exists between two variables. The graphing calculator does not provide an indication of how well the function fits the data in terms of r, since a quadratic function cannot be expressed as a linear function. Figure 32 Now Work PROBLEM 25 4.4 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Translate the following sentence into a mathematical equation: The total revenue R from selling x hot dogs is $3 times the number of hot dogs sold. (pp. 134–140) 2. Use a graphing utility to find the line of best fit for the following data: (pp. 284–287) x 3 5 5 6 7 8 y 10 13 12 15 16 19 308 CHAPTER 4 Linear and Quadratic Functions Applications and Extensions 3. Maximizing Revenue The price p (in dollars) and the quantity x sold of a certain product obey the demand equation 1 x + 100 6 Find a model that expresses the revenue R as a function of x. (Remember, R = xp.) What is the domain of R? What is the revenue if 200 units are sold? What quantity x maximizes revenue? What is the maximum revenue? What price should the company charge to maximize revenue? p = - (a) (b) (c) (d) (e) 4. Maximizing Revenue The price p (in dollars) and the quantity x sold of a certain product obey the demand equation 1 x + 100 3 Find a model that expresses the revenue R as a function of x. What is the domain of R? What is the revenue if 100 units are sold? What quantity x maximizes revenue? What is the maximum revenue? What price should the company charge to maximize revenue? 9. Enclosing the Most Area with a Fence A farmer with 4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed? (See the figure.) x x p = - (a) (b) (c) (d) (e) 5. Maximizing Revenue The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = - 5p + 100 0 6 p … 20 (a) Express the revenue R as a function of x. (b) What is the revenue if 15 units are sold? (c) What quantity x maximizes revenue? What is the maximum revenue? (d) What price should the company charge to maximize revenue? (e) What price should the company charge to earn at least $480 in revenue? 6. Maximizing Revenue The price p (in dollars) and the quantity x sold of a certain product obey the demand equation x = - 20p + 500 0 6 p … 25 (a) Express the revenue R as a function of x. (b) What is the revenue if 20 units are sold? (c) What quantity x maximizes revenue? What is the maximum revenue? (d) What price should the company charge to maximize revenue? (e) What price should the company charge to earn at least $3000 in revenue? 7. Enclosing a Rectangular Field David has 400 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width w of the rectangle. (b) For what value of w is the area largest? (c) What is the maximum area? 8. Enclosing a Rectangular Field Beth has 3000 feet of fencing available to enclose a rectangular field. (a) Express the area A of the rectangle as a function of x, where x is the length of the rectangle. (b) For what value of x is the area largest? (c) What is the maximum area? 4000 2x 10. Enclosing the Most Area with a Fence A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed? 11. Analyzing the Motion of a Projectile A projectile is fired from a cliff 200 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 50 feet per second. The height h of the projectile above the water is modeled by h1x2 = - 32x2 + x + 200 1502 2 where x is the horizontal distance of the projectile from the face of the cliff. (a) At what horizontal distance from the face of the cliff is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the face of the cliff will the projectile strike the water? (d) Using a graphing utility, graph the function h, 0 … x … 200. (e) Use a graphing utility to verify the solutions found in parts (b) and (c). (f) When the height of the projectile is 100 feet above the water, how far is it from the cliff? 12. Analyzing the Motion of a Projectile A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 100 feet per second. The height h of the projectile is modeled by h1x2 = - 32x2 + x 11002 2 where x is the horizontal distance of the projectile from the firing point. (a) At what horizontal distance from the firing point is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the firing point will the projectile strike the ground? (d) Using a graphing utility, graph the function h, 0 … x … 350. SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data 309 (e) Use a graphing utility to verify the results obtained in parts (b) and (c). (f) When the height of the projectile is 50 feet above the ground, how far has it traveled horizontally? 13. Suspension Bridge A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 100 meters from the center. (Assume that the road is level.) 14. Architecture A parabolic arch has a span of 120 feet and a maximum height of 25 feet. Choose suitable rectangular coordinate axes and find the equation of the parabola. Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center. 15. Constructing Rain Gutters A rain gutter is to be made of aluminum sheets that are 12 inches wide by turning up the edges 90°. See the illustration. (a) What depth will provide maximum cross-sectional area and hence allow the most water to flow? (b) What depths will allow at least 16 square inches of water to flow? 18. Architecture A special window has the shape of a rectangle surmounted by an equilateral triangle. See the figure. If the perimeter of the window is 16 feet, what dimensions will admit the most light? 13 2 [Hint: Area of an equilateral triangle = a bx , where x 4 is the length of a side of the triangle.] x x x 19. Chemical Reactions A self-catalytic chemical reaction results in the formation of a compound that causes the formation ratio to increase. If the reaction rate V is modeled by x x 12 12 2x in. x x V 1x2 = kx1a - x2, 0 … x … a where k is a positive constant, a is the initial amount of the compound, and x is the variable amount of the compound, for what value of x is the reaction rate a maximum? 20. Calculus: Simpson’s Rule The figure shows the graph of y = ax2 + bx + c. Suppose that the points 1 - h, y0 2, 10, y1 2, and 1h, y2 2 are on the graph. It can be shown that the area enclosed by the parabola, the x-axis, and the lines x = - h and x = h is Area = h 12ah2 + 6c2 3 Show that this area may also be given by 16. Norman Windows A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. See the figure. If the perimeter of the window is 20 feet, what dimensions will admit the most light (maximize the area)? [Hint: Circumference of a circle = 2pr; area of a circle = pr 2, where r is the radius of the circle.] Area = h 1y + 4y1 + y2 2 3 0 y (0, y1) (h, y2) (h, y0) h h x 21. Use the result obtained in Problem 20 to find the area enclosed by f 1x2 = - 5x2 + 8, the x-axis, and the lines x = - 1 and x = 1. 22. Use the result obtained in Problem 20 to find the area enclosed by f 1x2 = 2x2 + 8, the x-axis, and the lines x = - 2 and x = 2. 17. Constructing a Stadium A track-and-field playing area is in the shape of a rectangle with semicircles at each end. See the figure (top, right). The inside perimeter of the track is to be 1500 meters. What should the dimensions of the rectangle be so that the area of the rectangle is a maximum? 23. Use the result obtained in Problem 20 to find the area enclosed by f 1x2 = x2 + 3x + 5, the x-axis, and the lines x = - 4 and x = 4. 24. Use the result obtained in Problem 20 to find the area enclosed by f 1x2 = - x2 + x + 4, the x-axis, and the lines x = - 1 and x = 1. 310 CHAPTER 4 Linear and Quadratic Functions 25. Life Cycle Hypothesis An individual’s income varies with his or her age. The following table shows the median income I of males of different age groups within the United States for 2012. For each age group, let the class midpoint represent the independent variable, x. For the class “65 years and older,” we will assume that the class midpoint is 69.5. Class Midpoint, x Median Income, I 15–24 years 19.5 $10,869 25–34 years 29.5 $34,113 35–44 years 39.5 $45,225 45–54 years 49.5 $46,466 55–64 years 59.5 $42,176 65 years and older 69.5 $27,612 Age Source: U.S. Census Bureau (a) Use a graphing utility to draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) Use a graphing utility to find the quadratic function of best fit that models the relation between age and median income. (c) Use the function found in part (b) to determine the age at which an individual can expect to earn the most income. (d) Use the function found in part (b) to predict the peak income earned. (e) With a graphing utility, graph the quadratic function of best fit on the scatter diagram. 26. Height of a Ball A shot-putter throws a ball at an inclination of 45° to the horizontal. The following data represent the height of the ball h, in feet, at the instant that it has traveled x feet horizontally. Distance, x Height, h 20 25 40 40 60 55 80 65 100 71 120 77 140 77 160 75 180 71 200 64 (a) Use a graphing utility to draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) Use a graphing utility to find the quadratic function of best fit that models the relation between distance and height. (c) Use the function found in part (b) to determine how far the ball will travel before it reaches its maximum height. (d) Use the function found in part (b) to find the maximum height of the ball. (e) With a graphing utility, graph the quadratic function of best fit on the scatter diagram. Mixed Practice 27. Which Model? The following data represent the square footage and rents (dollars per month) for apartments in the La Jolla area of San Diego, California. Square Footage, x Rent per Month, R 520 $1525 621 $1750 718 $1785 753 $1850 850 $1900 968 $2130 1020 $2180 Source: apartments.com, 2014 (a) Using a graphing utility, draw a scatter diagram of the data treating square footage as the independent variable. What type of relation appears to exist between square footage and rent? (b) Based on your response to part (a), find either a linear or a quadratic model that describes the relation between square footage and rent. (c) Use your model to predict the rent for an apartment in San Diego that is 875 square feet. 28. Which Model? An engineer collects the following data showing the speed s of a Toyota Camry and its average miles per gallon, M. Speed, s Miles per Gallon, M 30 18 35 20 40 23 40 25 45 25 50 28 55 30 60 29 65 26 65 25 70 25 (a) Using a graphing utility, draw a scatter diagram of the data, treating speed as the independent variable. What type of relation appears to exist between speed and miles per gallon? SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data (b) Based on your response to part (a), find either a linear model or a quadratic model that describes the relation between speed and miles per gallon. (c) Use your model to predict the miles per gallon for a Camry that is traveling 63 miles per hour. 29. Which Model? The following data represent the birth rate (births per 1000 population) for women whose age is a, in 2012. Age, a 30. Which Model? A cricket makes a chirping noise by sliding its wings together rapidly. Perhaps you have noticed that the number of chirps seems to increase with the temperature. The following data list the temperature (in degrees Fahrenheit) and the number of chirps per second for the striped ground cricket. Birth Rate, B 16 14.1 19 51.4 22 83.1 27 106.5 32 97.3 37 48.3 42 10.4 Temperature (ºF), x Chirps per Second, C 88.6 20.0 93.3 19.8 80.6 17.1 69.7 14.7 69.4 15.4 79.6 15.0 80.6 16.0 76.3 14.4 75.2 15.5 Source: Pierce, George W. The Songs of Insects. Cambridge, MA Harvard University Press, 1949, pp. 12 – 21 Source: National Vital Statistics System, 2013 (a) Using a graphing utility, draw a scatter diagram of the data, treating age as the independent variable. What type of relation appears to exist between age and birth rate? (b) Based on your response to part (a), find either a linear or a quadratic model that describes the relation between age and birth rate. (c) Use your model to predict the birth rate for 35-yearold women. 311 (a) Using a graphing utility, draw a scatter diagram of the data, treating temperature as the independent variable. What type of relation appears to exist between temperature and chirps per second? (b) Based on your response to part (a), find either a linear or a quadratic model that best describes the relation between temperature and chirps per second. (c) Use your model to predict the chirps per second if the temperature is 80°F. Explaining Concepts: Discussion and Writing 31. Refer to Example 1 in this section. Notice that if the price charged for the calculators is $0 or $140, then the revenue is $0. It is easy to explain why revenue would be $0 if the price charged were $0, but how can revenue be $0 if the price charged is $140? Retain Your Knowledge Problems 32–35 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 32. Express as a complex number: 2- 225 33. Find the distance between the points P1 = (4, - 7) and P2 = ( - 1, 5). 34. Find the equation of the circle with center (– 6, 0) and radius r = 27. 35. Solve: 5x2 + 8x - 3 = 0 ‘Are You Prepared?’ Answers 1. R = 3x 2. y = 1.7826x + 4.0652 312 CHAPTER 4 Linear and Quadratic Functions 4.5 Inequalities Involving Quadratic Functions PREPARING FOR THIS SECTION Before getting started, review the following: r Solve Inequalities (Section 1.5, pp. 123–126) r Use Interval Notation (Section 1.5, pp. 120–121) Now Work the ‘Are You Prepared?’ problems on page 314. OBJECTIVE 1 Solve Inequalities Involving a Quadratic Function (p. 312) 1 Solve Inequalities Involving a Quadratic Function In this section we solve inequalities that involve quadratic functions. We will accomplish this by using their graphs. For example, to solve the inequality ax2 + bx + c 7 0 a ≠ 0 graph the function f1x2 = ax2 + bx + c and, from the graph, determine where it is above the x-axis—that is, where f1x2 7 0. To solve the inequality ax2 + bx + c 6 0, a ≠ 0, graph the function f 1x2 = ax2 + bx + c and determine where the graph is below the x-axis. If the inequality is not strict, include the x-intercepts, if any, in the solution. Solving an Inequality EX AMPLE 1 Solve the inequality x2 - 4x - 12 … 0 and graph the solution set. Solution y 8 Graph the function f 1x2 = x2 - 4x - 12. f102 = - 12 y-intercept: x-intercepts (if any): 4 (6, 0) (–2, 0) –4 8 x 4 x - 4x - 12 = 0 2 1x - 62 1x + 22 = 0 x - 6 = 0 or x + 2 = 0 x = 6 or x = -2 –4 Evaluate f at 0. Solve f1x2 = 0. Factor. Apply the Zero-Product Property. –8 The y-intercept is - 12; the x-intercepts are - 2 and 6. b -4 = = 2. Because f122 = - 16, the vertex is The vertex is at x = 2a 2 at 12, - 162 . –12 (0, –12) –16 (2, –16) Figure 33 f (x) = x2 - 4x - 12 –4 –2 0 2 4 6 8 x See Figure 33 for the graph. The graph is below the x-axis for - 2 6 x 6 6. Because the original inequality is not strict, include the x-intercepts. The solution set is 5 x 0 - 2 … x … 66 or, using interval notation, 3 - 2, 64 . See Figure 34 for the graph of the solution set. r Figure 34 Now Work EX AMPLE 2 PROBLEM 9 Solving an Inequality Solve the inequality 2x2 6 x + 10 and graph the solution set. Solution Option 1 Rearrange the inequality so that 0 is on the right side. 2x2 6 x + 10 Subtract x + 10 from both sides. 2x2 - x - 10 6 0 This inequality is equivalent to the original inequality. SECTION 4.5 Inequalities Involving Quadratic Functions Next graph the function f1x2 = 2x2 - x - 10 to find where f1x2 6 0. y 4 f102 = - 10 y-intercept: 2 (2, 0) 2 2 x-intercepts (if any): 4 x 6 ( 1–4 , 10.125) Figure 35 f (x) = 2x2 - x - 10 g(x) x 10 y 12 –– ) ( 5–2 , 25 2 10 Option 2 If f1x2 = 2x2 and g1x2 = x + 10, then the inequality to be solved is f1x2 6 g1x2. Graph the functions f1x2 = 2x2 and g1x2 = x + 10. See Figure 36. The graphs intersect where f1x2 = g1x2. Then 2x2 - x - 10 = 0 f(x) 2x 2 12x - 52 1x + 22 = 0 4 2 4 x = x Figure 36 2 0 2 4 EXAM PLE 3 y 3 PROBLEMS 5 AND 13 Solving an Inequality Solve the inequality x2 + x + 1 7 0 and graph the solution set. (1, 3) Solution 2 (0, 1) –2 –1 1 2 x ( 1–2, 3–4) Figure 38 f 1x2 = x2 + x + 1 2 Figure 39 x = -2 or r Now Work 4 5 2 Apply the Zero-Product Property. 5 25 The graphs intersect at the points 1 - 2, 82 and a , b . To solve f 1x2 6 g1x2, find 2 2 where the graph of f is below the graph of g. This happens between the points of 5 intersection. Because the inequality is strict, the solution set is e x ` - 2 6 x 6 f or, 2 5 using interval notation, a - 2, b . 2 See Figure 37 for the graph of the solution set. Figure 37 (–1, 1) Factor. 2x - 5 = 0 or x + 2 = 0 2 4 f1x2 = g1x2 2x2 = x + 10 2 Apply the Zero-Product Property. 5 The y-intercept is - 10; the x-intercepts are - 2 and . 2 b -1 1 1 The vertex is at x = = = . Because f a b = - 10.125, the vertex 2a 4 4 4 1 is a , - 10.125b . See Figure 35 for the graph. 4 5 The graph is below the x-axis 1f1x2 6 02 between x = - 2 and x = . Because 2 5 the inequality is strict, the solution set is e x ` - 2 6 x 6 f or, using interval 2 5 notation, a - 2, b . 2 8 6 4 Solve f1x2 = 0. Factor. 2x - 5 = 0 or x + 2 = 0 5 x = or x = - 2 2 2 (2, 8) Evaluate f at 0. 2x - x - 10 = 0 12x - 52 1x + 22 = 0 ( 5–2 , 0) 4 10 313 0 2 Graph the x-intercepts b x = = 2a and 1 - 1, 12 function f 1x2 = x2 + x + 1. The y-intercept is 1; there are no (Do you see why? Check the discriminant). The vertex is at 1 1 3 1 3 - . Since f a - b = , the vertex is at a - , b . The points 11, 32 2 2 4 2 4 are also on the graph. See Figure 38. The graph of f lies above the x-axis for all x. The solution set is the set of all real numbers, or ( - q , q ). See Figure 39. r 4 Now Work PROBLEM 17 314 CHAPTER 4 Linear and Quadratic Functions 4.5 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 2. Write 1 - 2, 7] using inequality notation. (pp. 120–121) 1. Solve the inequality - 3x - 2 6 7 (pp. 123–126) Skill Building In Problems 3–6, use the figure to solve each inequality. 3. y y f(x) 4. y 5 5. (1.5, 5) (2, 8) 3 (2, 0) y 6. y f(x) 8 3 3 x 4 3 (3, 12) (4, 0) 2 2 4 x 4 (1, 2) 4 5 3 x (1, 3) 12 4 (3, 12) y f(x) x y g(x) 4 (a) f 1x2 7 0 (b) f 1x2 … 0 y g(x) 6 (2, 8) (2, 0) (1, 0) y y g(x) (a) g1x2 Ú f 1x2 (b) f 1x2 7 g1x2 (a) g1x2 6 0 (b) g1x2 Ú 0 (a) f 1x2 6 g1x2 (b) f 1x2 Ú g1x2 In Problems 7–22, solve each inequality. 7. 11. 15. 19. x2 - 3x - 10 6 0 x2 - 9 6 0 2x2 6 5x + 3 4x2 + 9 6 6x 8. 12. 16. 20. x2 + 3x - 10 7 0 x2 - 1 6 0 6x2 6 6 + 5x 25x2 + 16 6 40x 9. 13. 17. 21. x2 - 4x 7 0 x2 + x 7 12 x2 - x + 1 … 0 61x2 - 12 7 5x 10. 14. 18. 22. x2 + 8x x2 + 7x x2 + 2x 212x2 - 7 0 6 - 12 + 4 7 0 3x2 7 - 9 Mixed Practice 23. What is the domain of the function f 1x2 = 2x2 - 16? In Problems 25–32, use the given functions f and g. (a) Solve f 1x2 = 0. (e) Solve g1x2 … 0. 25. f 1x2 = x2 - 1 g1x2 = 3x + 3 29. f 1x2 = x2 - 4 g1x2 = - x2 + 4 (b) Solve g1x2 = 0. 24. What is the domain of the function f 1x2 = 2x - 3x2? (c) Solve f 1x2 = g1x2. (d) Solve f 1x2 7 0. 27. f 1x2 = - x2 + 1 g1x2 = 4x + 1 28. f 1x2 = - x2 + 4 g1x2 = - x - 2 (f) Solve f 1x2 7 g1x2. (g) Solve f 1x2 Ú 1. 30. f 1x2 = x2 - 2x + 1 g1x2 = - x2 + 1 31. f 1x2 = x2 - x - 2 g1x2 = x2 + x - 2 26. f 1x2 = - x2 + 3 g1x2 = - 3x + 3 32. f 1x2 = - x2 - x + 1 g1x2 = - x2 + x + 6 Applications and Extensions 33. Physics A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s 1t2 = 80t - 16t 2. 96 ft s 80t 16t 2 (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 96 feet above the ground? 34. Physics A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s 1t2 = 96t - 16t 2. (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 128 feet above the ground? 35. Revenue Suppose that the manufacturer of a gas clothes dryer has found that when the unit price is p dollars, the revenue R (in dollars) is R 1p2 = - 4p2 + 4000p Chapter Review (a) At what prices p is revenue zero? (b) For what range of prices will revenue exceed $800,000? (a) If the round must clear a hill 200 meters high at a distance of 2000 meters in front of the howitzer, what c values are permitted in the trajectory equation? (b) If the goal in part (a) is to hit a target on the ground 75 kilometers away, is it possible to do so? If so, for what values of c? If not, what is the maximum distance the round will travel? Source: www.answers.com 36. Revenue The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. The revenue R, in dollars, is given by R 1p2 = - 1 2 p + 1900p 2 (a) At what prices p is revenue zero? (b) For what range of prices will revenue exceed $1,200,000? 37. Artillery A projectile fired from the point (0, 0) at an angle to the positive x-axis has a trajectory given by g x 2 y = cx - 11 + c 2 2 a b a b 2 v where x = y = v = g = horizontal distance in meters height in meters initial muzzle velocity in meters per second (m/sec) acceleration due to gravity = 9.81 meters per second squared (m/sec2) c 7 0 is a constant determined by the angle of elevation. A howitzer fires an artillery round with a muzzle velocity of 897 m/sec. 315 38. Runaway Car Using Hooke’s Law, we can show that the work done in compressing a spring a distance of x feet from its 1 at-rest position is W = kx2, where k is a stiffness constant 2 depending on the spring. It can also be shown that the work done by a body in motion before it comes to rest is given w 2 ∼ v , where w = weight of the object (in lbs), by W = 2g g = acceleration due to gravity (32.2 ft/sec2), and v = object’s velocity (in ft/sec). A parking garage has a spring shock absorber at the end of a ramp to stop runaway cars. The spring has a stiffness constant k = 9450 lb/ft and must be able to stop a 4000-lb car traveling at 25 mph. What is the least compression required of the spring? Express your answer using feet to the nearest tenth. ∼ [Hint: Solve W 7 W, x Ú 0]. Source: www.sciforums.com Explaining Concepts: Discussion and Writing 39. Show that the inequality 1x - 42 2 … 0 has exactly one solution. 40. Show that the inequality 1x - 22 7 0 has one real number that is not a solution. 2 2 41. Explain why the inequality x + x + 1 7 0 has all real numbers as the solution set. 42. Explain why the inequality x2 - x + 1 6 0 has the empty set as the solution set. 43. Explain the circumstances under which the x-intercepts of the graph of a quadratic function are included in the solution set of a quadratic inequality. Retain Your Knowledge Problems 44–47 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 2 44. Determine the domain of f(x) = 210 - 2x. 46. Consider the linear function f(x) = x - 6. 3 -x is even, 45. Determine algebraically whether f(x) = 2 (a) Find the intercepts of the graph of f. x + 9 odd, or neither. (b) Graph f. 47. Multiply (6 - 5i)(4 - i). Write the answer in the form a + bi. ‘Are You Prepared?’ Answers 1. 5x 0 x 7 - 36 or 1 - 3, q 2 2. - 2 6 x … 7 Chapter Review Things to Know Linear function (p. 274) f 1x2 = mx + b Average rate of change = m The graph is a line with slope m and y-intercept b. 316 CHAPTER 4 Linear and Quadratic Functions Quadratic function (pp. 291–295) f 1x2 = ax2 + bx + c, a ≠ 0 The graph is a parabola that opens up if a 7 0 and opens down if a 6 0. Vertex: a - b b , f a- b b 2a 2a Axis of symmetry: x = y-intercept: f 102 = c b 2a x-intercept(s): If any, found by finding the real solutions of the equation ax2 + bx + c = 0 Objectives Section 4.1 You should be able to… Examples Review Exercises Graph linear functions (p. 274) Use average rate of change to identify linear functions (p. 274) Determine whether a linear function is increasing, decreasing, or constant (p. 277) Build linear models from verbal descriptions (p. 278) 1 2 3 4, 5 1(a)–3(a), 1(c)–3(c) 1(b)–3(b), 4, 5 1(d)–3(d) 21 Draw and interpret scatter diagrams (p. 284) Distinguish between linear and nonlinear relations (p. 285) Use a graphing utility to find the line of best fit (p. 286) 1 2 4 29(a), 30(a) 29(b), 30(a) 29(c) Graph a quadratic function using transformations (p. 292) Identify the vertex and axis of symmetry of a quadratic function (p. 294) Graph a quadratic function using its vertex, axis, and intercepts (p. 294) Find a quadratic function given its vertex and one other point (p. 297) Find the maximum or minimum value of a quadratic function (p. 298) 1 2 3–5 6 7 6–8 9–13 9–13 19, 20 14–16, 22–27 2 Build quadratic models from verbal descriptions (p. 302) Build quadratic models from data (p. 306) 1–4 5 22–28 30 3 Solve inequalities involving a quadratic function (p. 312) 1–3 17, 18 1 2 3 4 4.2 1 2 3 4.3 1 2 3 4 5 4.4 1 4.5 Review Exercises In Problems 1–3: (a) Determine the slope and y-intercept of each linear function. (b) Find the average rate of change of each function. (c) Graph each function. Label the intercepts. (d) Determine whether the function is increasing, decreasing, or constant. 4 1. f 1x2 = 2x - 5 2. h1x2 = x - 6 5 3. G1x2 = 4 In Problems 4 and 5, determine whether the function is linear or nonlinear. If the function is linear, state its slope. 4. x y = f(x) -1 5. x y = g(x) -2 -1 -3 0 3 0 4 1 8 1 7 2 13 2 6 3 18 3 1 In Problems 6–8, graph each quadratic function using transformations (shifting, compressing, stretching, and/or reflecting). 6. f 1x2 = 1x + 12 2 - 4 7. f 1x2 = - 1x - 42 2 8. f 1x2 = - 31x + 22 2 + 1 Chapter Review 317 In Problems 9–13, (a) graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of the function. (c) Determine where the function is increasing and where it is decreasing. 1 9. f 1x2 = 1x - 22 2 + 2 10. f 1x2 = x2 - 16 11. f 1x2 = - 4x2 + 4x 4 9 12. f 1x2 = x2 + 3x + 1 13. f 1x2 = 3x2 + 4x - 1 2 In Problems 14–16, determine whether the given quadratic function has a maximum value or a minimum value, and then find the value. 15. f 1x2 = - x2 + 8x - 4 14. f 1x2 = 3x2 - 6x + 4 16. f 1x2 = - 3x2 + 12x + 4 In Problems 17 and 18, solve each quadratic inequality. 17. x2 + 6x - 16 6 0 18. 3x2 Ú 14x + 5 In Problems 19 and 20, find the quadratic function for which: 19. Vertex is (2, −4); y-intercept is −16 20. Vertex is 1 - 1, 22; contains the point 11, 62 21. Sales Commissions Bill was just offered a sales position for a computer company. His salary would be $25,000 per year plus 1% of his total annual sales. (a) Find a linear function that relates Bill’s annual salary, S, to his total annual sales, x. (b) If Bill’s total annual sales were $1,000,000, what would be Bill’s salary? (c) What would Bill have to sell to earn $100,000? (d) Determine the sales required of Bill for his salary to exceed $150,000. 22. Demand Equation The price p (in dollars) and the quantity x sold of a certain product obey the demand equation p = - 1 x + 150 10 0 … x … 1500 (a) Express the revenue R as a function of x. (b) What is the revenue if 100 units are sold? (c) What quantity x maximizes revenue? What is the maximum revenue? (d) What price should the company charge to maximize revenue? 25. Architecture A special window in the shape of a rectangle with semicircles at each end is to be constructed so that the outside perimeter is 100 feet. See the illustration. Find the dimensions of the rectangle that maximizes its area. 26. Minimizing Marginal Cost Callaway Golf Company has determined that the marginal cost C of manufacturing x Big Bertha golf clubs may be expressed by the quadratic function C 1x2 = 4.9x2 - 617.4x + 19,600 (a) How many clubs should be manufactured to minimize the marginal cost? (b) At this level of production, what is the marginal cost? 27. Maximizing Area A rectangle has one vertex on the line y = 10 - x, x 7 0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Express the area A of the rectangle as a function of x. Find the largest area A that can be enclosed by the rectangle. 28. Parabolic Arch Bridge A horizontal bridge is in the shape of a parabolic arch. Given the information shown in the figure, what is the height h of the arch 2 feet from shore? 23. Landscaping A landscape engineer has 200 feet of border to enclose a rectangular pond. What dimensions will result in the largest pond? 24. Enclosing the Most Area with a Fence A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. See the figure. What is the largest area that can be enclosed? 10 ft h 2 ft 20 ft 29. Bone Length Research performed at NASA, led by Dr. Emily R. Morey-Holton, measured the lengths of the right humerus and right tibia in 11 rats that were sent to space on Spacelab Life Sciences 2. The data on page 318 were collected. (a) Draw a scatter diagram of the data, treating length of the right humerus as the independent variable. (b) Based on the scatter diagram, do you think that there is a linear relation between the length of the right humerus and the length of the right tibia? 318 CHAPTER 4 Linear and Quadratic Functions (c) Use a graphing utility to find the line of best fit relating length of the right humerus and length of the right tibia. (d) Predict the length of the right tibia on a rat whose right humerus is 26.5 millimeters (mm). (a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. Advertising Expenditures ($1000s) Total Revenue ($1000s) 20 6101 6222 Right Humerus (mm), x Right Tibia (mm), y 22 25 6350 24.80 36.05 25 6378 24.59 35.57 27 6453 24.59 35.57 28 6423 24.29 34.58 29 6360 23.81 34.20 31 6231 24.87 34.73 25.90 37.38 26.11 37.96 26.63 37.46 26.31 37.75 26.84 38.50 (b) The quadratic function of best fit to these data is R 1A2 = - 7.76A2 + 411.88A + 942.72 Source: NASA Life Sciences Data Archive 30. Advertising A small manufacturing firm collected the following data on advertising expenditures A (in thousands of dollars) and total revenue R (in thousands of dollars). Chapter Test Use this function to determine the optimal level of advertising. (c) Use the function to predict the total revenue when the optimal level of advertising is spent. (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) Use a graphing utility to draw a scatter diagram of the data, and then graph the quadratic function of best fit on the scatter diagram. The Chapter Test Prep Videos are step-by-step solutions available in , or on this text’s Channel. Flip back to the Resources for Success page for a link to this text’s YouTube channel. 1. Consider the linear function f 1x2 = - 4x + 3: (a) Find the slope and y-intercept. (b) What is the average rate of change of f ? (c) Determine whether f is increasing, decreasing, or constant. (d) Graph f. In Problems 2 and 3, find the intercepts, if any, of each quadratic function. 2. f 1x2 = 3x2 - 2x - 8 3. G1x2 = - 2x2 + 4x + 1 4. Given that f 1x2 = x2 + 3x and g1x2 = 5x + 3, solve f 1x2 = g1x2. Graph each function and label the points of intersection. 5. Graph f 1x2 = 1x - 32 2 - 2 using transformations. 6. Consider the quadratic function f 1x2 = 3x2 - 12x + 4: (a) Determine whether the graph opens up or down. (b) Determine the vertex. (c) Determine the axis of symmetry. (d) Determine the intercepts. (e) Use the information from parts (a)–(d) to graph f. 7. Determine whether f 1x2 = - 2x2 + 12x + 3 has a maximum or a minimum. Then find the maximum or minimum value. 8. Solve x2 - 10x + 24 Ú 0. 9. RV Rental The weekly rental cost of a 20-foot recreational vehicle is $129.50 plus $0.15 per mile. (a) Find a linear function that expresses the cost C as a function of miles driven m. (b) What is the rental cost if 860 miles are driven? (c) How many miles were driven if the rental cost is $213.80? Cumulative Review 319 Cumulative Review 1. Find the distance between the points P = 1 - 1, 32 and Q = 14, - 22. Find the midpoint of the line segment P to Q. 2. Which of the following points are on the graph of y = x3 - 3x + 1? (a) 1 - 2, - 12 (b) (2, 3) (c) (3, 1) 3. Solve the inequality 5x + 3 Ú 0 and graph the solution set. 4. Find the equation of the line containing the points 1 - 1, 42 and 12, - 22. Express your answer in slope–intercept form and graph the line. 5. Find the equation of the line perpendicular to the line y = 2x + 1 and containing the point (3, 5). Express your answer in slope–intercept form and graph both lines. 6. Graph the equation x2 + y2 - 4x + 8y - 5 = 0. 7. Does the following relation represent a function? 5 1 - 3, 82, 11, 32, 12, 52, 13, 82 6 . 8. For the function f defined by f 1x2 = x2 - 4x + 1, find: (a) f 122 (b) f 1x2 + f 122 (c) f 1 - x2 (d) - f 1x2 f 1x + h2 - f 1x2 (e) f 1x + 22 (f) , h ≠ 0 h 3z - 1 . 9. Find the domain of h1z2 = 6z - 7 10. Is the following graph the graph of a function? x2 even, odd, or neither? 2x + 1 13. Approximate the local maximum values and local minimum values of f 1x2 = x3 - 5x + 1 on 1 - 4, 42. Determine where the function is increasing and where it is decreasing. 12. Is the function f 1x2 = 14. If f 1x2 = 3x + 5 and g1x2 = 2x + 1: (a) Solve f 1x2 = g1x2. (b) Solve f 1x2 7 g1x2. 15. Consider the graph below of the function f. (a) Find the domain and the range of f. (b) Find the intercepts. (c) Is the graph of f symmetric with respect to the x-axis, the y-axis, or the origin? (d) Find f(2). (e) For what value(s) of x is f 1x2 = 3? (f) Solve f 1x2 6 0. (g) Graph y = f 1x2 + 2. (h) Graph y = f 1 - x2. (i) Graph y = 2f 1x2. (j) Is f even, odd, or neither? (k) Find the interval(s) on which f is increasing. y 4 (4, 3) (2, 1) y 5 x 11. Consider the function f 1x2 = (4, 3) x . x + 4 1 (a) Is the point a1, b on the graph of f ? 4 (b) If x = - 2, what is f (x)? What point is on the graph of f ? (c) If f 1x2 = 2, what is x? What point is on the graph of f ? (2, 1) (1, 0) 4 (1, 0) (0, 1) 5 x 320 CHAPTER 4 Linear and Quadratic Functions Chapter Projects variable and the percentage change in the stock you chose as the dependent variable. The easiest way to draw a scatter diagram in Excel is to place the two columns of data next to each other (for example, have the percentage change in the S&P500 in column F and the percentage change in the stock you chose in column G). Then highlight the data and select the Scatter Diagram icon under Insert. Comment on the type of relation that appears to exist between the two variables. Internet-based Project I. The Beta of a Stock You want to invest in the stock market but are not sure which stock to purchase. Information is the key to making an informed investment decision. One piece of information that many stock analysts use is the beta of the stock. Go to Wikipedia (http://en.wikipedia.org/wiki/ Beta_%28finance%29) and research what beta measures and what it represents. 1. 3. Finding beta. To find beta requires that we find the line of best fit using least-squares regression. The easiest approach is to click inside the scatter diagram. Select the Chart Elements icon ( + ). Check the box for Trendline, select the arrow to the right, and choose More Options. Select Linear and check the box for Display Equation on chart. The line of best fit appears on the scatter diagram. See below. Approximating the beta of a stock. Choose a wellknown company such as Google or Coca-Cola. Go to a website such as Yahoo! Finance (http://finance. yahoo.com/) and find the weekly closing price of the company’s stock for the past year. Then find the closing price of the Standard & Poor’s 500 (S&P500) for the same time period. To get the historical prices in Yahoo! Finance, select Historical Prices from the left menu. Choose the appropriate time period. Select Weekly and Get Prices. Finally, select Download to Spreadsheet. Repeat this for the S&P500, and copy the data into the same spreadsheet. Finally, rearrange the data in chronological order. Be sure to expand the selection to sort all the data. Now, using the adjusted close price, compute the percentage change in price for each P1 - Po week, using the formula % change = . For Po example, if week 1 price is in cell D1 and week 2 price is D2 - D1 in cell D2, then % change = . Repeat this for D1 the S&P500 data. 2. Using Excel to draw a scatter diagram. Treat the percentage change in the S&P500 as the independent y 0.08 0.06 0.04 0.02 0.1 0.05 O 0.05 0.1 x 0.02 0.04 0.06 y 0.9046x 0.0024 R 2 0.4887 Series 1 Linear (series 1) The line of best fit for this data is y = 0.9046x + 0.0024. You may click on Chart Title or either axis title and insert the appropriate names. The beta is the slope of the line of best fit, 0.9046. We interpret this by saying, “If the S&P500 increases by 1%, then this stock will increase by 0.9%, on average.” Find the beta of your stock and provide an interpretation. NOTE: Another way to use Excel to find the line of best fit requires using the Data Analysis Tool Pack under add-ins. The following projects are available on the Instructor’s Resource Center (IRC): II. Cannons A battery commander uses the weight of a missile, its initial velocity, and the position of its gun to determine where the missile will travel. III. First and Second Differences Finite differences provide a numerical method that is used to estimate the graph of an unknown function. IV. CBL Experiment Computer simulation is used to study the physical properties of a bouncing ball. 5 Polynomial and Rational Functions Day Length Day length is the length of time each day from the moment the upper limb of the sun’s disk appears above the horizon during sunrise to the moment when the upper limb disappears below the horizon during sunset. The length of a day depends on the day of the year as well as the latitude of the location. Latitude gives the location of a point on Earth north or south of the equator. In the Internet Project at the end of this chapter, we use information from the chapter to investigate the relation between day length and latitude for a specific day of the year. —See the Internet-based Chapter Project I— A Look Back In Chapter 3, we began our discussion of functions. We defined domain, range, and independent and dependent variables, found the value of a function, and graphed functions. We continued our study of functions by listing the properties that a function might have, such as being even or odd, and created a library of functions, naming key functions and listing their properties, including their graphs. In Chapter 4, we discussed linear functions and quadratic functions, which belong to the class of polynomial functions. A Look Ahead In this chapter, we look at two general classes of functions, polynomial functions and rational functions, and examine their properties. Polynomial functions are arguably the simplest expressions in algebra. For this reason, they are often used to approximate other, more complicated functions. Rational functions are ratios of polynomial functions. Outline 5.1 5.2 5.3 5.4 5.5 5.6 Polynomial Functions and Models Properties of Rational Functions The Graph of a Rational Function Polynomial and Rational Inequalities The Real Zeros of a Polynomial Function Complex Zeros; Fundamental Theorem of Algebra Chapter Review Chapter Test Cumulative Review Chapter Projects 321 322 CHAPTER 5 Polynomial and Rational Functions 5.1 Polynomial Functions and Models PREPARING FOR THIS SECTION Before getting started, review the following: r Graphing Techniques: Transformations (Section 3.5, pp. 247–256) r Intercepts (Section 2.2, pp. 159–160) r Polynomials (Chapter R, Section R.4, pp. 39–47) r Using a Graphing Utility to Approximate Local Maxima and Local Minima (Section 3.3, p. 229) r Intercepts of a Function (Section 3.2, pp. 215–217) Now Work the ‘Are You Prepared?’ problems on page 338. OBJECTIVES 1 Identify Polynomial Functions and Their Degree (p. 322) 2 Graph Polynomial Functions Using Transformations (p. 326) 3 Identify the Real Zeros of a Polynomial Function and Their Multiplicity (p. 327) 4 Analyze the Graph of a Polynomial Function (p. 334) 5 Build Cubic Models from Data (p. 336) 1 Identify Polynomial Functions and Their Degree In Chapter 4, we studied the linear function f1x2 = mx + b, which can be written as f1x2 = a1x + a0 and the quadratic function f1x2 = ax2 + bx + c, a ≠ 0, which can be written as f1x2 = a2x2 + a1x + a0 a2 ≠ 0 Each of these functions is an example of a polynomial function. DEFINITION A polynomial function in one variable is a function of the form f 1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0 (1) where an , an - 1 , c, a1 , a0 are constants, called the coefficients of the polynomial, n Ú 0 is an integer, and x is the variable. If an ≠ 0, it is called the leading coefficient, and n is the degree of the polynomial. The domain of a polynomial function is the set of all real numbers. In Words A polynomial function is a sum of monomials. EX AMPLE 1 The monomials that make up a polynomial are called its terms. If an ≠ 0, anxn is called the leading term; a0 is called the constant term. If all of the coefficients are 0, the polynomial is called the zero polynomial, which has no degree. Polynomials are usually written in standard form, beginning with the nonzero term of highest degree and continuing with terms in descending order according to degree. If a power of x is missing, it is because its coefficient is zero. Polynomial functions are among the simplest expressions in algebra. They are easy to evaluate: only addition and repeated multiplication are required. Because of this, they are often used to approximate other, more complicated functions. In this section, we investigate properties of this important class of functions. Identifying Polynomial Functions Determine which of the following are polynomial functions. For those that are, state the degree; for those that are not, tell why not. Write each polynomial in standard form, and then identify the leading term and the constant term. 1 (a) p 1x2 = 5x3 - x2 - 9 (b) f1x2 = x + 2 - 3x4 (c) g1x2 = 1x 4 x2 - 2 (d) h 1x2 = 3 (e) G1x2 = 8 (f) H1x2 = - 2x3 1x - 12 2 x - 1 SECTION 5.1 Polynomial Functions and Models Solution 323 (a) p is a polynomial function of degree 3, and it is already in standard form. The leading term is 5x3, and the constant term is - 9. (b) f is a polynomial function of degree 4. Its standard form is f1x2 = - 3x4 + x + 2. The leading term is - 3x4, and the constant term is 2. 1 (c) g is not a polynomial function because g1x2 = 1x = x2, so the variable x is 1 raised to the power, which is not a nonnegative integer. 2 (d) h is not a polynomial function. It is the ratio of two distinct polynomials, and the polynomial in the denominator is of positive degree. (e) G is a nonzero constant polynomial function so it is of degree 0. The polynomial is in standard form. The leading term and constant term are both 8. (f) H 1x2 = - 2x3 1x - 12 2 = - 2x3 1x2 - 2x + 12 = - 2x5 + 4x4 - 2x3. So, H is a polynomial function of degree 5. Because H1x2 = - 2x5 + 4x4 - 2x3, the leading term is - 2x5. Since no constant term is shown, the constant term is 0. Do you see a way to find the degree of H without multiplying it out? r Now Work 17 PROBLEMS AND 21 We have already discussed in detail polynomial functions of degrees 0, 1, and 2. See Table 1 for a summary of the properties of the graphs of these polynomial functions. Table 1 Degree Form Name Graph No degree f(x) = 0 Zero function The x-axis 0 f(x) = a0 , a0 ≠ 0 Constant function Horizontal line with y-intercept a0 1 f(x) = a1 x + a0 , a1 ≠ 0 Linear function Nonvertical, nonhorizontal line with slope a1 and y-intercept a0 2 f(x) = a2 x2 + a1 x + a0 , a2 ≠ 0 Quadratic function Parabola: graph opens up if a2 7 0; graph opens down if a2 6 0 One objective of this section is to analyze the graph of a polynomial function. If you take a course in calculus, you will learn that the graph of every polynomial function is both smooth and continuous. By smooth, we mean that the graph contains no sharp corners or cusps; by continuous, we mean that the graph has no gaps or holes and can be drawn without lifting your pencil from the paper. See Figures 1(a) and (b). y y Cusp Corner Gap Hole x Figure 1 (a) Graph of a polynomial function: smooth, continuous x (b) Cannot be the graph of a polynomial function Power Functions We begin the analysis of the graph of a polynomial function by discussing power functions, a special kind of polynomial function. DEFINITION In Words A power function is defined by a single monomial. A power function of degree n is a monomial function of the form f 1x2 = axn where a is a real number, a ≠ 0, and n 7 0 is an integer. (2) 324 CHAPTER 5 Polynomial and Rational Functions Examples of power functions are f1x2 = 3x f1x2 = - 5x2 degree 1 degree 2 f1x2 = 8x3 f1x2 = - 5x4 degree 3 degree 4 The graph of a power function of degree 1, f1x2 = ax, is a straight line, with slope a, that passes through the origin. The graph of a power function of degree 2, f1x2 = ax2, is a parabola, with vertex at the origin, that opens up if a 7 0 and opens down if a 6 0. If we know how to graph a power function of the form f1x2 = xn, a compression or stretch and, perhaps, a reflection about the x-axis will enable us to obtain the graph of g1x2 = axn. Consequently, we shall concentrate on graphing power functions of the form f1x2 = xn. We begin with power functions of even degree of the form f1x2 = xn, n Ú 2 and n even. The domain of f is the set of all real numbers, and the range is the set of nonnegative real numbers. Such a power function is an even function. (Do you see why?). Its graph is symmetric with respect to the y-axis. Its graph always contains the origin and the points 1 - 1, 12 and 11, 12 . If n = 2, the graph is the familiar parabola y = x2 that opens up, with vertex at the origin. If n Ú 4, the graph of f1x2 = xn, n even, will be closer to the x-axis than the parabola y = x2 if - 1 6 x 6 1, x ≠ 0, and farther from the x-axis than the parabola y = x2 if x 6 - 1 or if x 7 1. Figure 2(a) illustrates this conclusion. Figure 2(b) shows the graphs of y = x4 and y = x8 for comparison. f (x ) = x n n≥4 n even y 4 2 2 (1, 1) (– 1, 1) (0, 0) 3 x (1, 1) (0, 0) –3 (a) Figure 2 y=x4 y 4 (– 1, 1) –3 y=x8 y = x2 3 x (b) Figure 2 shows that as n increases, the graph of f1x2 = xn, n Ú 2 and n even, tends to flatten out near the origin and is steeper when x is far from 0. For large n, it may appear that the graph coincides with the x-axis near the origin, but it does not; the graph actually touches the x-axis only at the origin (see Table 2). Also, for large n, it may appear that for x 6 - 1 or for x 7 1 the graph is vertical, but it is not; it is only increasing very rapidly in these intervals. If the graphs were enlarged many times, these distinctions would be clear. x = 0.1 x = 0.3 f (x) = x 8 10 -8 0.0000656 0.0039063 f (x) = x 20 10 -20 3.487 # 10 0.000001 f (x) = x 40 10 -40 1.216 # 10-21 Table 2 x = 0.5 -11 9.095 # 10-13 Seeing the Concept Graph Y1 = x4, Y2 = x8, and Y3 = x12 using the viewing rectangle - 2 … x … 2, - 4 … y … 16. Then graph each again using the viewing rectangle - 1 … x … 1, 0 … y … 1. See Figure 3. TRACE along one of the graphs to confirm that for x close to 0 the graph is above the x-axis and that for x 7 0 the graph is increasing. SECTION 5.1 Polynomial Functions and Models 16 325 1 Y3 5 Y1 5 x 4 x 12 Y2 5 x 8 Y1 5 x 4 Y2 5 x 8 22 2 21 24 Figure 3 Y3 5 x 12 1 0 (b) (a) Properties of Power Functions, f (x) = xn, n Is a Positive Even Integer y 3 1. f is an even function, so its graph is symmetric with respect to the y-axis. 2. The domain is the set of all real numbers. The range is the set of nonnegative real numbers. 3. The graph always contains the points 1 - 1, 12 , 10, 02 , and 11, 12 . 4. As the exponent n increases in magnitude, the graph is steeper when x 6 - 1 or x 7 1; but for x near the origin, the graph tends to flatten out and lie closer to the x-axis. y = x3 y=xn n≥5 n odd (1, 1) –3 3 (0, 0) x (–1, –1) –3 Figure 4 y 3 y = x9 y = x5 Now we consider power functions of odd degree of the form f1x2 = xn, n Ú 3 and n odd. The domain and the range of f are the set of real numbers. Such a power function is an odd function. (Do you see why?). Its graph is symmetric with respect to the origin. Its graph always contains the origin and the points 1 - 1, - 12 and 11, 12 . The graph of f 1x2 = xn when n = 3 has been shown several times and is repeated in Figure 4. If n Ú 5, the graph of f1x2 = xn, n odd, will be closer to the x-axis than that of y = x3 if - 1 6 x 6 1 and farther from the x-axis than that of y = x3 if x 6 - 1 or if x 7 1. Figure 4 illustrates this conclusion. Figure 5 shows the graphs of y = x5 and y = x9 for further comparison. It appears that each graph coincides with the x-axis near the origin, but it does not; each graph actually crosses the x-axis at the origin. Also, it appears that as x increases the graphs become vertical, but they do not; each graph is just increasing very rapidly. (1, 1) Seeing the Concept –3 (0, 0) 3 x (–1, –1) Graph Y1 = x3, Y2 = x7, and Y3 = x11 using the viewing rectangle - 2 … x … 2, - 16 … y … 16. Then graph each again using the viewing rectangle - 1 … x … 1, - 1 … y … 1. See Figure 6. TRACE along one of the graphs to confirm that the graph is increasing and crosses the x-axis at the origin. –3 16 1 Figure 5 Y1 5 x 3 Y2 5 x 7 Y3 5 x 11 Y2 5 x 7 Y1 5 x 3 22 2 216 Figure 6 (a) 21 Y3 5 x 11 1 21 (b) 326 CHAPTER 5 Polynomial and Rational Functions To summarize: Properties of Power Functions, f (x) = xn, n Is a Positive Odd Integer f is an odd function, so its graph is symmetric with respect to the origin. The domain and the range are the set of all real numbers. The graph always contains the points 1 - 1, - 12 , 10, 02 , and 11, 12 . As the exponent n increases in magnitude, the graph is steeper when x 6 - 1 or x 7 1; but for x near the origin, the graph tends to flatten out and lie closer to the x-axis. 1. 2. 3. 4. 2 Graph Polynomial Functions Using Transformations The methods of shifting, compression, stretching, and reflection (studied in Section 3.5), when used with the facts just presented, will enable us to graph polynomial functions that are transformations of power functions. EX AMPLE 2 Graphing a Polynomial Function Using Transformations Graph: f1x2 = 1 - x5 Solution It is helpful to rewrite f as f1x2 = - x5 + 1. Figure 7 shows the required stages. y y 2 2 (1, 1) 2 (0, 0) x 2 2 (1, 0) (0, 0) x 2 x 2 2 2 Multiply by 1; reflect about x-axis Add 1; shift up 1 unit (b) y x 5 (a) y x 5 (c) y x 5 1 1 x 5 r Graphing a Polynomial Function Using Transformations Graph: f1x2 = Solution 2 (1, 1) 2 EX AMPLE 3 (0, 1) (1, 1) (1, 1) Figure 7 y 2 (1, 2) 1 1x - 12 4 2 Figure 8 shows the required stages. y 2 y 2 (1, 1) (1, 1) 2 (0, 0) 2 y 2 (0, 1) x 2 2 (2, 1) (1, 0) 2 x (0, 1–2) (2, 1–2) 2 (1, 0) 2 x 2 2 1– ; 2 Figure 8 (a) y x 4 Multiply by Replace x by x 1; shift right compression by 1 unit a factor of 1–2 (c) y (b) y ( x 1)4 Now Work PROBLEMS 29 AND 35 1– 2 ( x 1)4 r SECTION 5.1 Polynomial Functions and Models 327 3 Identify the Real Zeros of a Polynomial Function and Their Multiplicity Figure 9 shows the graph of a polynomial function with four x-intercepts. Notice that at the x-intercepts, the graph must either cross the x-axis or touch the x-axis. Consequently, between consecutive x-intercepts the graph is either above the x-axis or below the x-axis. y Above x-axis Above x -axis x Crosses x -axis Below x -axis Touches x -axis Crosses x -axis Below x -axis Figure 9 Graph of a polynomial function If a polynomial function f is factored completely, it is easy to locate the x-intercepts of the graph by solving the equation f1x2 = 0 and using the Zero-Product Property. For example, if f 1x2 = 1x - 12 2 1x + 32 , then the solutions of the equation f1x2 = 1x - 12 2 1x + 32 = 0 are identified as 1 and - 3. That is, f112 = 0 and f 1 - 32 = 0. DEFINITION If f is a function and r is a real number for which f 1r2 = 0, then r is called a real zero of f. As a consequence of this definition, the following statements are equivalent. 1. 2. 3. 4. r is a real zero of a polynomial function f. r is an x-intercept of the graph of f. x - r is a factor of f. r is a solution to the equation f1x2 = 0. So the real zeros of a polynomial function are the x-intercepts of its graph, and they are found by solving the equation f1x2 = 0. EXAMPL E 4 Finding a Polynomial Function from Its Zeros (a) Find a polynomial function of degree 3 whose zeros are - 3, 2, and 5. (b) Use a graphing utility to graph the polynomial found in part (a) to verify your result. Solution (a) If r is a real zero of a polynomial function f, then x - r is a factor of f. This means that x - 1 - 32 = x + 3, x - 2, and x - 5 are factors of f. As a result, any polynomial function of the form f1x2 = a1x + 32 1x - 22 1x - 52 where a is a nonzero real number, qualifies. The value of a causes a stretch, compression, or reflection, but it does not affect the x-intercepts of the graph. Do you know why? 328 CHAPTER 5 Polynomial and Rational Functions (b) We choose to graph f with a = 1. Then 40 24 f1x2 = 1x + 32 1x - 22 1x - 52 = x3 - 4x2 - 11x + 30 6 Figure 10 shows the graph of f. Notice that the x-intercepts are - 3, 2, and 5. r Seeing the Concept 250 Figure 10 f(x) = x3 - 4x2 - 11x + 30 Graph the function found in Example 4 for a = 2 and a = - 1. Does the value of a affect the zeros of f? How does the value of a affect the graph of f? Now Work PROBLEM 43 If the same factor x - r occurs more than once, r is called a repeated, or multiple, zero of f. More precisely, we have the following definition. DEFINITION EX AMPLE 5 If 1x - r2 m is a factor of a polynomial f and 1x - r2 m + 1 is not a factor of f, then r is called a zero of multiplicity m of f.* Identifying Zeros and Their Multiplicities For the polynomial f1x2 = 5x2 1x + 22 ax - In Words The multiplicity of a zero is the number of times its corresponding factor occurs. 1 4 b 2 r - 2 is a zero of multiplicity 1 because the exponent on the factor x + 2 is 1. r 0 is a zero of multiplicity 2 because the exponent on the factor x is 2. 1 1 r is a zero of multiplicity 4 because the exponent on the factor x - is 4. 2 2 Now Work PROBLEM r 57(a) Suppose that it is possible to completely factor a polynomial function and, as a result, locate all the x-intercepts of its graph (the real zeros of the function). These x-intercepts then divide the x-axis into open intervals and, on each such interval, the graph of the polynomial will be either above or below the x-axis over the entire interval. Let’s look at an example. EX AMPLE 6 Graphing a Polynomial Using Its x-Intercepts Consider the following polynomial: f 1x2 = 1x + 12 2 1x - 22 (a) Find the x- and y-intercepts of the graph of f. (b) Use the x-intercepts to find the intervals on which the graph of f is above the x-axis and the intervals on which the graph of f is below the x-axis. (c) Locate other points on the graph, and connect all the points plotted with a smooth, continuous curve. Solution (a) The y-intercept is f 102 = 10 + 12 2 10 - 22 = - 2. The x-intercepts satisfy the equation f1x2 = 1x + 12 2 1x - 22 = 0 from which we find 1x + 12 2 = 0 or x - 2 = 0 x = - 1 or x = 2 The x-intercepts are - 1 and 2. *Some books use the terms multiple root and root of multiplicity m. SECTION 5.1 Polynomial Functions and Models 329 (b) The two x-intercepts divide the x-axis into three intervals: 1 - q , - 12 1 - 1, 22 12, q 2 Since the graph of f crosses or touches the x-axis only at x = - 1 and x = 2, it follows that the graph of f is either above the x-axis 3 f 1x2 7 04 or below the x-axis 3 f1x2 6 04 on each of these three intervals. To see where the graph lies, we need only pick a number in each interval, evaluate f there, and see whether the value is positive (above the x-axis) or negative (below the x-axis). See Table 3. (c) In constructing Table 3, we obtained three additional points on the graph: 1 - 2, - 42, 11, - 42 and 13, 162 . Figure 11 illustrates these points, the intercepts, and a smooth, continuous curve (the graph of f ) connecting them. y –1 Table 3 2 x (3, 16) Interval ( - q , - 1) ( - 1, 2) (2, q ) Number chosen -2 1 3 Value of f f ( - 2) = - 4 f (1) = - 4 f (3) = 16 Location of graph Below x-axis Below x-axis Above x-axis Point on graph ( - 2, - 4) (1, - 4) (3, 16) 12 6 (– 1, 0) (2, 0) 3 –2 (– 2, – 4) x (0, – 2) –6 (1, –4) Figure 11 f 1x2 = 1x + 12 2 1x - 22 r Look again at Table 3. Since the graph of f1x2 = 1x + 12 2 1x - 22 is below the x-axis on both sides of - 1, the graph of f touches the x-axis at x = - 1, a zero of multiplicity 2. Since the graph of f is below the x-axis for x 6 2 and above the x-axis for x 7 2, the graph of f crosses the x-axis at x = 2, a zero of multiplicity 1. This suggests the following results: If r Is a Zero of Even Multiplicity Numerically: The sign of f 1x2 does not change from one side to the other side of r. Graphically: The graph of f touches the x-axis at r. If r Is a Zero of Odd Multiplicity Numerically: The sign of f1x2 changes from one side to the other side of r. Graphically: The graph of f crosses the x-axis at r. Now Work PROBLEM 57(b) Turning Points Look again at Figure 11 above. We cannot be sure just how low the graph actually goes between x = - 1 and x = 2. But we do know that somewhere in the interval 1 - 1, 22 the graph of f must change direction (from decreasing to increasing). The points at which a graph changes direction are called turning points.* Each turning point yields either a local maximum or a local minimum (see Section 3.3). The following result from calculus tells us the maximum number of turning points that the graph of a polynomial function can have. *Graphing utilities can be used to approximate turning points. For most polynomials, calculus is needed to find the exact turning points. 330 CHAPTER 5 Polynomial and Rational Functions THEOREM Turning Points If f is a polynomial function of degree n, then the graph of f has at most n - 1 turning points. If the graph of a polynomial function f has n - 1 turning points, then the degree of f is at least n. Based on the first part of the theorem, a polynomial function of degree 5 will have at most 5 - 1 = 4 turning points. Based on the second part of the theorem, if the graph of a polynomial function has three turning points, then the degree of the function must be at least 4. Exploration A graphing utility can be used to locate the turning points of a graph. Graph Y1 = (x + 1)2(x - 2). Use MINIMUM to find the location of the turning point for 0 6 x 6 2. See Figure 12. 0 2 26 Figure 12 Y1 = 1x + 12 2 1x - 22 Now Work EX AMPLE 7 PROBLEM 57(c) Identifying the Graph of a Polynomial Function Which of the graphs in Figure 13 could be the graph of a polynomial function? For those that could, list the real zeros and state the least degree the polynomial can have. For those that could not, say why not. y y y 2 2 2 –2 2 x –2 –2 2 –2 x –2 y 3 2 x 3 x –3 –2 –3 Figure 13 Solution (a) (b) (c) (d) (a) The graph in Figure 13(a) cannot be the graph of a polynomial function because of the gap that occurs at x = - 1. Remember, the graph of a polynomial function is continuous—no gaps or holes. (See Figure 1.) (b) The graph in Figure 13(b) could be the graph of a polynomial function because the graph is smooth and continuous. It has three real zeros: - 2, 1, and 2. Since the graph has two turning points, the degree of the polynomial function must be at least 3. (c) The graph in Figure 13(c) cannot be the graph of a polynomial function because of the cusp at x = 1. Remember, the graph of a polynomial function is smooth. (d) The graph in Figure 13(d) could be the graph of a polynomial function. It has two real zeros: - 2 and 1. Since the graph has three turning points, the degree of the polynomial function is at least 4. r Now Work PROBLEM 69 SECTION 5.1 Polynomial Functions and Models 331 End Behavior One last remark about Figure 11. For very large values of x, either positive or negative, the graph of f1x2 = 1x + 12 2 1x - 22 looks like the graph of y = x3. To see why, we write f in the form f 1x2 = 1x + 12 2 1x - 22 = x3 - 3x - 2 = x3 a1 - 3 2 - 3b 2 x x 3 2 Now, for large values of x, either positive or negative, the terms 2 and 3 are x x close to 0, so for large values of x, f1x2 = x3 - 3x - 2 = x3 a1 - 3 2 - 3 b ≈ x3 x2 x The behavior of the graph of a function for large values of x, either positive or negative, is referred to as its end behavior. THEOREM End Behavior For large values of x, either positive or negative, the graph of the polynomial function f1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0 In Words The end behavior of a polynomial function resembles that of its leading term. an ≠ 0 resembles the graph of the power function y = an xn For example, if f 1x2 = - 2x3 + 5x2 + x - 4, then the graph of f will behave like the graph of y = - 2x3 for very large values of x, either positive or negative. We can see that the graphs of f and y = - 2x3 “behave” the same by considering Table 4 and Figure 14. Table 4 x f(x) y = − 2x3 10 - 1,494 - 2,000 100 - 1,949,904 - 2,000,000 500 - 248,749,504 - 250,000,000 1,000 - 1,994,999,004 - 2,000,000,000 175 Y2 5 22x 3 1 5x 2 1 x 2 4 25 5 Y1 5 22x 3 2175 Figure 14 NOTE Infinity ( q ) and negative infinity ( - q ) are not numbers. Rather, they are symbols that represent unboundedness. ■ Notice that as x becomes a larger and larger positive number, the values of f become larger and larger negative numbers. When this happens, we say that f is unbounded in the negative direction. Rather than using words to describe the behavior of the graph of the function, we explain its behavior using notation. We can symbolize “the value of f becomes a larger and larger negative number as x becomes a larger and larger positive number” by writing f1x2 S - q as x S q (read “the values of f approach negative infinity as x approaches infinity”). In calculus, limits are used to convey these ideas. There we use the symbolism lim f1x2 = - q , read “the limit of f1x2 as x approaches infinity equals negative xS q infinity,” to mean that f1x2 S - q as x S q . 332 CHAPTER 5 Polynomial and Rational Functions When we say that the value of a limit equals infinity (or negative infinity), we mean that the values of the function are unbounded in the positive (or negative) direction and call the limit an infinite limit. When we discuss limits as x becomes unbounded in the negative direction or unbounded in the positive direction, we are discussing limits at infinity. Look back at Figures 2 and 4. Based on the preceding theorem and the previous discussion on power functions, the end behavior of a polynomial function can be of only four types. See Figure 15. f(x) as x ` 2` y f(x) as x ` ` y y x f(x) as x ` ` x 2` 2` f(x) as x f(x) as x E n ≥ 2 even; an < 0 D n ≥ 2 even; an > 0 2` ` f(x) as x ` 2` y x f(x) as x x 2` 2` f(x) as x F n ≥ 3 odd; an > 0 G n ≥ 3 odd; an < 0 2` ` Figure 15 End behavior of f(x) = anxn + an-1xn-1 + g + a1x + a0 For example, if f1x2 = - 2x4 + x3 + 4x2 - 7x + 1, the graph of f will resemble the graph of the power function y = - 2x4 for large 0 x 0 . The graph of f will behave like Figure 15(b) for large 0 x 0 . Now Work EX AMPLE 8 PROBLEM 57(d) Identifying the Graph of a Polynomial Function Which of the graphs in Figure 16 could be the graph of f(x) = x4 + ax3 + bx2 - 5x - 6 where a 7 0, b 7 0? y y x (a) y x (b) y x (c) x (d) Figure 16 Solution The y-intercept of f is f 102 = - 6. We can eliminate the graph in Figure 16(a), whose y-intercept is positive. We are not able to solve f1x2 = 0 to find the x-intercepts of f, so we move on to investigate the turning points of each graph. Since f is of degree 4, the graph of f has at most 3 turning points. We eliminate the graph in Figure 16(c) because that graph has 5 turning points. Now we look at end behavior. For large values of x, the graph of f will behave like the graph of y = x4. This eliminates the graph in Figure 16(d), whose end behavior is like the graph of y = - x4. SECTION 5.1 Polynomial Functions and Models 333 Only the graph in Figure 16(b) could be the graph of f 1x2 = x4 + ax3 + bx2 - 5x - 6 where a 7 0, b 7 0. EXAMPL E 9 r Writing a Polynomial Function from Its Graph Write a polynomial function whose graph is shown in Figure 17 (use the smallest degree possible). y 15 (– 1, 6) – 3 (– 2, 0) 9 (0, 0) (2, 0) 3 x –9 – 15 – 21 Figure 17 Solution The x-intercepts are - 2, 0, and 2. Therefore, the polynomial must have the factors 1x + 22 , x, and 1x - 22, respectively. There are three turning points, so the degree of the polynomial must be at least 4. The graph touches the x-axis at x = - 2, so - 2 must have an even multiplicity. The graph crosses the x-axis at x = 0 and x = 2, so 0 and 2 must have odd multiplicities. Using the smallest degree possible (1 for odd multiplicity and 2 for even multiplicity), we can write f1x2 = ax1x + 2221x - 22 All that remains is to find the leading coefficient, a. From Figure 17, the point 1 - 1, 62 must lie on the graph. 6 = a1 - 12 1 - 1 + 2221 - 1 - 22 6 = 3a 2 = a f (- 1) = 6 The polynomial function f1x2 = 2x1x + 2221x - 22 would have the graph in Figure 17. Check: Graph Y1 = 2x1x + 22 2 1x - 22 using a graphing utility to verify this result. r Now Work PROBLEMS 73 AND 77 SUMMARY Graph of a Polynomial Function f 1 x2 = an xn + an - 1 xn - 1 + g + a1x + a0 an 3 0 Degree of the polynomial function f : n y-intercept: f102 = a0. Graph is smooth and continuous. Maximum number of turning points: n - 1 At a zero of even multiplicity: The graph of f touches the x-axis. At a zero of odd multiplicity: The graph of f crosses the x-axis. Between zeros, the graph of f is either above or below the x-axis. End behavior: For large 0 x 0 , the graph of f behaves like the graph of y = an xn. 334 CHAPTER 5 Polynomial and Rational Functions 4 Analyze the Graph of a Polynomial Function How to Analyze the Graph of a Polynomial Function EX A MPL E 1 0 Analyze the factored form of the polynomial function f1x2 = 12x + 12 1x - 32 2. Step-by-Step Solution Expand the polynomial: f1x2 = 12x + 12 1x - 32 2 = 12x + 12 1x2 - 6x + 92 Step 1: Determine the end behavior of the graph of the function. = 2x3 - 12x2 + 18x + x2 - 6x + 9 Multiply. = 2x - 11x + 12x + 9 Combine like terms. 3 2 The polynomial function f is of degree 3. The graph of f behaves like y = 2x3 for large values of x . The y-intercept is f102 = 9. To find the x-intercepts, solve f 1x2 = 0. Step 2: Find the x- and y-intercepts of the graph of the function. f1x2 = 0 12x + 12 1x - 32 2 = 0 2x + 1 = 0 x = The x-intercepts are - 1 2 or 1x - 32 2 = 0 or x = 3 1 and 3. 2 Step 3: Determine the zeros of the function and their multiplicity. Use this information to determine whether the graph crosses or touches the x-axis at each x-intercept. 1 1 and 3. The zero - is a zero of multiplicity 1, so the graph of f 2 2 1 crosses the x-axis at x = - . The zero 3 is a zero of multiplicity 2, so the graph of f 2 touches the x-axis at x = 3. Step 4: Determine the maximum number of turning points on the graph of the function. Because the polynomial function is of degree 3 (Step 1), the graph of the function will have at most 3 - 1 = 2 turning points. Step 5: Put all the information from Steps 1 through 4 together to obtain the graph of f. To help establish the y-axis scale, find additional points on the graph on each side of any x-intercept. Figure 18(a) illustrates the information obtained from Steps 1 through 4. We evaluate f at - 1, 1, and 4 to help establish the scale on the y-axis. We find that f1 - 12 = - 16, f112 = 12, and f142 = 9, so we plot the points 1 - 1, - 162, 11, 122 , and (4, 9). The graph of f is given in Figure 18(b). The zeros of f are - y y End behavior: Resembles y = 2x 3 40 40 30 The graph y -intercept: 9 20 crosses the x-axis at 1 − 1– , 0 10 x = − –2 . ( 2 ) (3, 0) (0, 9) −2 −1 −10 1 Figure 18 3 20 ( − 1–2 , 4 The graph touches the x-axis at x = 3. −20 End behavior: Resembles y = 2x 3 2 −30 −40 30 5 x 0) 10 −2 −1 −10 (−1, −16) −20 (1, 12) 1 (4, 9) (3, 0) (0, 9) 2 3 4 5 x −30 −40 (a) (b) r SECTION 5.1 Polynomial Functions and Models 335 SUMMARY Analyzing the Graph of a Polynomial Function STEP 1: Determine the end behavior of the graph of the function. STEP 2: Find the x- and y-intercepts of the graph of the function. STEP 3: Determine the zeros of the function and their multiplicity. Use this information to determine whether the graph crosses or touches the x-axis at each x-intercept. STEP 4: Determine the maximum number of turning points on the graph of the function. STEP 5: Use the information in Steps 1 through 4 to draw a complete graph of the function. To help establish the y-axis scale, find additional points on the graph on each side of any x-intercept. Now Work PROBLEM 81 For polynomial functions that have noninteger coefficients and for polynomials that are not easily factored, we use a graphing utility early in the analysis. This is because the amount of information that can be obtained from algebraic analysis is limited. EX AM PL E 1 1 How to Use a Graphing Utility to Analyze the Graph of a Polynomial Function Analyze the graph of the polynomial function f1x2 = x3 + 2.48x2 - 4.3155x + 2.484406 Step-by-Step Solution Step 1: Determine the end behavior of the graph of the function. The polynomial function f is of degree 3. The graph of f behaves like y = x3 for large values of x . Step 2: Graph the function using a graphing utility. See Figure 19 for the graph of f. 15 25 3 210 Figure 19 f(y1) = x3 + 2.48x2 - 4.3155x + 2.484406 Step 3: Use a graphing utility to approximate the x- and y-intercepts of the graph. The y-intercept is f102 = 2.484406. In Example 10, the polynomial function was factored, so it was easy to find the x-intercepts algebraically. However, it is not readily apparent how to factor f in this example. Therefore, we use a graphing utility’s ZERO (or ROOT or SOLVE) feature and find the lone x-intercept to be - 3.79, rounded to two decimal places. 336 CHAPTER 5 Polynomial and Rational Functions Step 4: Use a graphing utility to create a TABLE to find points on the graph around each x-intercept. Table 5 below shows values of x on each side of the x-intercept. The points 1 - 4, - 4.572 and 1 - 2, 13.042 are on the graph. Table 5 Step 5: Approximate the turning points of the graph. From the graph of f shown in Figure 19, we can see that f has two turning points. Using MAXIMUM reveals one turning point is at ( - 2.28, 13.36), rounded to two decimal places. Using MINIMUM shows that the other turning point is at (0.63, 1), rounded to two decimal places. Step 6: Use the information in Steps 1 through 5 to draw a complete graph of the function by hand. Figure 20 shows a graph of f drawn by hand using the information in Steps 1 through 5. y (–2.28, 13.36) 12 (–2, 13.04) End behavior: Resembles y = x 3 6 (0, 2.484406) –5 Figure 20 Step 7: Find the domain and the range of the function. Step 8: Use the graph to determine where the function is increasing and where it is decreasing. End behavior: Resembles y = x 3 (–3.79, 0) (–4, –4.57) –6 2x (0.63, 1) The domain and the range of f are the set of all real numbers. Based on the graph, f is increasing on the intervals 1 - q , - 2.282 and 10.63, q 2 . Also, f is decreasing on the interval 1 - 2.28, 0.632 . r SUMMARY Using a Graphing Utility to Analyze the Graph of a Polynomial Function STEP 1: STEP 2: STEP 3: STEP 4: STEP 5: STEP 6: STEP 7: STEP 8: Determine the end behavior of the graph of the function. Graph the function using a graphing utility. Use a graphing utility to approximate the x- and y-intercepts of the graph. Use a graphing utility to create a TABLE to find points on the graph around each x-intercept. Approximate the turning points of the graph. Use the information in Steps 1 through 5 to draw a complete graph of the function by hand. Find the domain and the range of the function. Use the graph to determine where the function is increasing and where it is decreasing. Now Work PROBLEM 99 5 Build Cubic Models from Data In Section 4.2 we found the line of best fit from data, and in Section 4.4 we found the quadratic function of best fit. It is also possible to find polynomial functions of SECTION 5.1 Polynomial Functions and Models 337 best fit. However, most statisticians do not recommend finding polynomials of best fit of degree higher than 3. Data that follow a cubic relation should look like Figure 21(a) or (b). y 5 ax 3 1 bx 2 1 cx 1 d, a . 0 y 5 ax 3 1 bx 2 1 cx 1 d, a , 0 D E Figure 21 Cubic relation A Cubic Function of Best Fit EX AM PL E 1 2 The data in Table 6 represent the weekly cost C (in thousands of dollars) of printing x thousand textbooks. Table 6 Number x of Textbooks, (thousands) (a) Draw a scatter diagram of the data using x as the independent variable and C as the dependent variable. Comment on the type of relation that may exist between the two variables x and C. (b) Using a graphing utility, find the cubic function of best fit C = C 1x2 that models the relation between number of texts and cost. (c) Graph the cubic function of best fit on your scatter diagram. (d) Use the function found in part (b) to predict the cost of printing 22 thousand texts per week. Cost, C ($1000s) 0 100 5 128.1 10 144 13 153.5 17 161.2 Solution 18 162.6 20 166.3 23 178.9 25 190.2 27 221.8 (a) Figure 22 shows the scatter diagram. A cubic relation may exist between the two variables. (b) Upon executing the CUBIC REGression program, we obtain the results shown in Figure 23. The output that the utility provides shows us the equation y = ax3 + bx2 + cx + d. The cubic function of best fit to the data is C 1x2 = 0.0155x3 - 0.5951x2 + 9.1502x + 98.4327. (c) Figure 24 shows the graph of the cubic function of best fit on the scatter diagram. The function fits the data reasonably well. 250 22 250 22 30 0 Figure 22 Figure 23 0 30 Figure 24 (d) Evaluate the function C 1x2 at x = 22. C 1222 = 0.01551222 3 - 0.59511222 2 + 9.15021222 + 98.4327 ≈ 176.8 The model predicts that the cost of printing 22 thousand textbooks in a week will be 176.8 thousand dollars—that is, $176,800. r 338 CHAPTER 5 Polynomial and Rational Functions 5.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. The intercepts of the equation 9x2 + 4y = 36 are (pp. 159–160) . 2. Is the expression 4x3 - 3.6x2 - 12 a polynomial? If so, what is its degree? (pp. 39–47) 3. To graph y = x - 4, you would shift the graph of y = x a distance of units. (pp. 247–256) 2 2 5. True or False The x-intercepts of the graph of a function y = f 1x2 are the real solutions of the equation f 1x2 = 0. (pp. 215–217) 6. If g152 = 0, what point is on the graph of g? What is the corresponding x-intercept of the graph of g? (pp. 215–217) 4. Use a graphing utility to approximate (rounded to two decimal places) the local maximum value and local minimum value of f 1x2 = x3 - 2x2 - 4x + 5, for - 3 6 x 6 3. (p. 229) Concepts and Vocabulary 12. The graph of the function f 1x2 = 3x4 - x3 + 5x2 - 2x - 7 will behave like the graph of for large values of x . 7. The graph of every polynomial function is both and . 8. If r is a real zero of even multiplicity of a polynomial function f, (crosses/touches) the x-axis at r. then the graph of f 9. The graphs of power functions of the form f 1x2 = xn, where n is an even integer, always contain the points , , and . 10. If r is a solution to the equation f 1x2 = 0, name three additional statements that can be made about f and r, assuming f is a polynomial function. 11. The points at which a graph changes direction (from increasing to decreasing or decreasing to increasing) are called . 13. If f 1x2 = - 2x5 + x3 - 5x2 + 7, then and lim f 1x2 = . lim f 1x2 = xS -q x Sq 14. Explain what the notation lim f 1x2 = - q means. x Sq of a zero is the number of times its corresponding 15. The factor occurs. (a) degree (b) multiplicity (c) turning point (d) limit 16. The graph of y = 5x6 - 3x4 + 2x - 9 has at most how many turning points? (a) - 9 (b) 14 (c) 6 (d) 5 Skill Building In Problems 17–28, determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell why not. Write each polynomial in standard form. Then identify the leading term and the constant term. 1 - x2 17. f 1x2 = 4x + x3 18. f 1x2 = 5x2 + 4x4 19. g1x2 = 2 20. h1x2 = 3 - 1 x 2 23. g1x2 = x3/2 - x2 + 2 26. F 1x2 = x2 - 5 x3 21. f 1x2 = 1 - 22. f 1x2 = x1x - 12 1 x 24. h1x2 = 1x1 1x - 12 25. F 1x2 = 5x4 - px3 + 27. G1x2 = 21x - 12 2 1x2 + 12 28. G1x2 = - 3x2 1x + 22 3 1 2 In Problems 29–42, use transformations of the graph of y = x4 or y = x5 to graph each function. 29. f 1x2 = 1x + 12 4 30. f 1x2 = 1x - 22 5 31. f 1x2 = x5 - 3 32. f 1x2 = x4 + 2 33. f 1x2 = 34. f 1x2 = 3x5 35. f 1x2 = - x5 36. f 1x2 = - x4 37. f 1x2 = 1x - 12 5 + 2 38. f 1x2 = 1x + 22 4 - 3 39. f 1x2 = 21x + 12 4 + 1 40. f 1x2 = 41. f 1x2 = 4 - 1x - 22 5 42. f 1x2 = 3 - 1x + 22 4 1 4 x 2 1 1x - 12 5 - 2 2 339 SECTION 5.1 Polynomial Functions and Models In Problems 43–50, form a polynomial function whose real zeros and degree are given. Answers will vary depending on the choice of a leading coefficient. 43. Zeros: - 1, 1, 3; degree 3 44. Zeros: - 2, 2, 3; degree 3 45. Zeros: - 3, 0, 4; degree 3 46. Zeros: - 4, 0, 2; degree 3 47. Zeros: - 4, - 1, 2, 3; degree 4 48. Zeros: - 3, - 1, 2, 5; degree 4 49. Zeros: - 1, multiplicity 1; 3, multiplicity 2; degree 3 50. Zeros: - 2, multiplicity 2; 4, multiplicity 1; degree 3 In Problems 51–56, find the polynomial function with the given zeros whose graph passes through the given point. 51. Zeros: - 3, 1, 4 Point: 16, 1802 52. Zeros: - 2, 0, 2 Point: 1 - 4, 162 53. Zeros: - 1, 0, 2, 4 54. Zeros: - 5, - 1, 2, 6 55. Zeros: - 1 (multiplicity 2), 1 (multiplicity 2) Point: 1 - 2, 452 56. Zeros: - 1 (multiplicity 2), 0, 3 (multiplicity 2) Point: 11,- 482 5 Point: ¢ , 15≤ 2 1 Point: ¢ , 63≤ 2 In Problems 57–68, for each polynomial function: (a) List each real zero and its multiplicity. (b) Determine whether the graph crosses or touches the x-axis at each x-intercept. (c) Determine the maximum number of turning points on the graph. (d) Determine the end behavior; that is, find the power function that the graph of f resembles for large values of x . 57. f 1x2 = 31x - 72 1x + 32 2 58. f 1x2 = 41x + 42 1x + 32 3 60. f 1x2 = 21x - 32 1x2 + 42 3 61. f 1x2 = - 2ax + 63. f 1x2 = 1x - 52 3 1x + 42 2 64. f 1x2 = 1x + 132 2 1x - 22 4 65. f 1x2 = 31x2 + 82 1x2 + 92 2 66. f 1x2 = - 21x2 + 32 3 67. f 1x2 = - 2x2 1x2 - 22 68. f 1x2 = 4x1x2 - 32 59. f 1x2 = 41x2 + 12 1x - 22 3 1 2 b 1x + 42 3 2 62. f 1x2 = ax - 1 2 b 1x - 12 3 3 In Problems 69–72, identify which of the graphs could be the graph of a polynomial function. For those that could, list the real zeros and state the least degree the polynomial can have. For those that could not, say why not. 69. 70. y 4 71. y 4 72. y y 4 2 2 2 –2 –4 4 x 2 –2 –4 2 –2 –2 –2 –4 –4 2 4 x 2 x –2 4 2 2 4 x 2 In Problems 73–76, construct a polynomial function that might have the given graph. (More than one answer may be possible.) 73. 74. y y 0 1 2 75. x 0 1 2 76. y y 2 2 1 1 x –2 –1 1 2 3 x –2 –1 1 –1 –1 –2 –2 2 3 x 340 CHAPTER 5 Polynomial and Rational Functions In Problems 77–80, write a polynomial function whose graph is shown (use the smallest degree possible). 77. 78. y 10 79. y 14 80. y 72 (3, 8) (– 2, 16) 6x –6 6 x –6 6x –6 –2 x 2 (2, – 50) (1, –8) –14 y 21 – 10 – 15 – 72 In Problems 81–98, analyze each polynomial function by following Steps 1 through 5 on page 335. 81. f 1x2 = x2 1x - 32 82. f 1x2 = x1x + 22 2 83. f 1x2 = 1x + 42 2 11 - x2 84. f 1x2 = 1x - 12 1x + 32 2 85. f 1x2 = - 21x + 22 1x - 22 3 1 86. f 1x2 = - 1x + 42 1x - 12 3 2 87. f 1x2 = 1x + 12 1x - 22 1x + 42 88. f 1x2 = 1x - 12 1x + 42 1x - 32 89. f 1x2 = x2 1x - 22 1x + 22 90. f 1x2 = x2 1x - 32 1x + 42 91. f 1x2 = 1x + 12 2 1x - 22 2 92. f 1x2 = 1x - 42 2 1x + 22 2 93. f 1x2 = x2 1x + 32 1x + 12 94. f 1x2 = x2 1x - 32 1x - 12 95. f 1x2 = 5x1x2 - 42 1x + 32 96. f 1x2 = 1x - 22 2 1x + 22 1x + 42 97. f 1x2 = x2 1x - 22 1x2 + 32 98. f 1x2 = x2 1x2 + 12 1x + 42 In Problems 99–106, analyze each polynomial function f by following Steps 1 through 8 on page 336. 99. f 1x2 = x3 + 0.2x2 - 1.5876x - 0.31752 100. f 1x2 = x3 - 0.8x2 - 4.6656x + 3.73248 101. f 1x2 = x3 + 2.56x2 - 3.31x + 0.89 102. f 1x2 = x3 - 2.91x2 - 7.668x - 3.8151 103. f 1x2 = x4 - 2.5x2 + 0.5625 104. f1x2 = x4 - 18.5x2 + 50.2619 105. f 1x2 = 2x4 - px3 + 15x - 4 106. f 1x2 = - 1.2x4 + 0.5x2 - 13x + 2 Mixed Practice In Problems 107–114, analyze each polynomial function by following Steps 1 through 5 on page 335. [Hint: You will need to first factor the polynomial]. 107. f 1x2 = 4x - x3 108. f 1x2 = x - x3 109. f 1x2 = x3 + x2 - 12x 110. f 1x2 = x3 + 2x2 - 8x 111. f 1x2 = 2x4 + 12x3 - 8x2 - 48x 112. f1x2 = 4x3 + 10x2 - 4x - 10 113. f 1x2 = - x5 - x4 + x3 + x2 114. f 1x2 = - x5 + 5x4 + 4x3 - 20x2 In Problems 115–118, construct a polynomial function f with the given characteristics. 116. Zeros: - 4, - 1, 2; degree 3; y-intercept: 16 115. Zeros: - 3, 1, 4; degree 3; y-intercept: 36 117. Zeros: - 5(multiplicity 2); 2 (multiplicity 1); 4 (multiplicity 1); degree 4; contains the point (3, 128) 118. Zeros: - 4 (multiplicity 1); 0 (multiplicity 3); 2 (multiplicity 1); degree 5; contains the point 1 - 2, 642 119. G1x2 = 1x + 32 2 1x - 22 (a) Identify the x-intercepts of the graph of G. (b) What are the x-intercepts of the graph y = G1x + 32 ? 120. h1x2 = 1x + 22 1x - 42 3 (a) Identify the x-intercepts of the graph of h. (b) What are the x-intercepts of the graph y = h1x - 22 ? of of SECTION 5.1 Polynomial Functions and Models 341 Applications and Extensions 121. Hurricanes In 2012, Hurricane Sandy struck the East Coast of the United States, killing 147 people and causing an estimated $75 billion in damage. With a gale diameter of about 1000 miles, it was the largest ever to form over the Atlantic Basin. The accompanying data represent the number of major hurricane strikes in the Atlantic Basin (category 3, 4, or 5) each decade from 1921 to 2010. Decade, x Major Hurricanes Striking Atlantic Basin, H 1921–1930, 1 17 1931–1940, 2 16 1941–1950, 3 29 1951–1960, 4 33 1961–1970, 5 27 1971–1980, 6 16 1981–1990, 7 16 1991–2000, 8 27 2001–2010, 9 33 Year, t Percent below Poverty Level, p 1990, 1 13.5 2001, 12 11.7 1991, 2 14.2 2002, 13 12.1 1992, 3 14.8 2003, 14 12.5 1993, 4 15.1 2004, 15 12.7 1994, 5 14.5 2005, 16 12.6 1995, 6 13.8 2006, 17 12.3 1996, 7 13.7 2007, 18 12.5 1997, 8 13.3 2008, 19 13.2 1998, 9 12.7 2009, 20 14.3 1999, 10 11.9 2010, 21 15.3 2000, 11 11.3 2011, 22 15.9 Source: U.S. Census Bureau 123. Temperature The following data represent the temperature T (°Fahrenheit) in Kansas City, Missouri, x hours after midnight on April 15, 2014. Source: National Oceanic & Atmospheric Administration Temperature (°F), T 3 36.1 6 32.0 9 39.0 12 46.2 15 52.0 18 55.0 21 52.0 24 48.9 (a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) Find the average rate of change in temperature from 9 am to 12 noon. (c) What is the average rate of change in temperature from 3 pm to 6 pm ? (d) Decide on a function of best fit to these data (linear, quadratic, or cubic) and use this function to predict the temperature at 5 pm. (e) With a graphing utility, draw a scatter diagram of the data and then graph the function of best fit on the scatter diagram. (f) Interpret the y-intercept. 124. Future Value of Money Suppose that you make deposits of $500 at the beginning of every year into an Individual Retirement Account (IRA) earning interest r (expressed as a decimal). At the beginning of the first year, the value of the account will be $500; at the beginning of the second year, the value of the account, will be +500 + +500r + Value of 1st deposit +500 = +50011 + r2 + +500 = 500r + 1000 6 122. Poverty Rates The following data represent the percentage of families in the United States whose income is below the poverty level. (a) With a graphing utility, draw a scatter diagram of the data. Comment on the type of relation that appears to exist between the two variables. (b) Decide on a function of best fit to these data (linear, quadratic, or cubic), and use this function to predict the percentage of U.S. families that were below the poverty level in 2012 (t = 23). Compare your prediction to the actual value of 15.9. (c) Draw the function of best fit on the scatter diagram drawn in part (a). Hours after Midnight, x Source: The Weather Underground e (a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) Use a graphing utility to find the cubic function of best fit that models the relation between decade and number of major hurricanes. (c) Use the model found in part (b) to predict the number of major hurricanes that struck the Atlantic Basin between 1961 and 1970. (d) With a graphing utility, draw a scatter diagram of the data and then graph the cubic function of best fit on the scatter diagram. (e) Concern has risen about the increase in the number and intensity of hurricanes, but some scientists believe this is just a natural fluctuation that could last another decade or two. Use your model to predict the number of major hurricanes that will strike the Atlantic Basin between 2011 and 2020. Is your result reasonable? Percent below Poverty Level, p Year, t Value of 2nd deposit 342 CHAPTER 5 Polynomial and Rational Functions (a) Verify that the value of the account at the beginning of the third year is T 1r2 = 500r 2 + 1500r + 1500. (b) The account value at the beginning of the fourth year is F1r2 = 500r 3 + 2000r 2 + 3000r + 2000. If the annual rate of interest is 5% = 0.05, what will be the value of the account at the beginning of the fourth year? 125. A Geometric Series In calculus, you will learn that certain functions can be approximated by polynomial functions. We will explore one such function now. (a) Using a graphing utility, create a table of values with 1 Y1 = f 1x2 = and Y2 = g2 1x2 = 1 + x + x2 + x3 1 - x for - 1 6 x 6 1 with ∆Tbl = 0.1. (b) Using a graphing utility, create a table of values with 1 and Y1 = f 1x2 = 1-x Y2 = g3 1x2 = 1 + x + x2 + x3 + x4 for - 1 6 x 6 1 with ∆Tbl = 0.1. (c) Using a graphing utility, create a table of values with 1 Y1 = f 1x2 = and 1 - x Y2 = g4 1x2 = 1 + x + x2 + x3 + x4 + x5 for - 1 6 x 6 1 with ∆Tbl = 0.1. (d) What do you notice about the values of the function as more terms are added to the polynomial? Are there some values of x for which the approximations are better? Explaining Concepts: Discussion and Writing 126. Can the graph of a polynomial function have no y-intercept? Can it have no x-intercepts? Explain. 132. The illustration shows the graph of a polynomial function. 127. Write a few paragraphs that provide a general strategy for graphing a polynomial function. Be sure to mention the following: degree, intercepts, end behavior, and turning points. y 128. Make up a polynomial that has the following characteristics: crosses the x-axis at - 1 and 4, touches the x-axis at 0 and 2, and is above the x-axis between 0 and 2. Give your polynomial to a fellow classmate and ask for a written critique. 129. Make up two polynomials, not of the same degree, with the following characteristics: crosses the x-axis at - 2, touches the x-axis at 1, and is above the x-axis between - 2 and 1. Give your polynomials to a fellow classmate and ask for a written critique. 130. The graph of a polynomial function is always smooth and continuous. Name a function studied earlier that is smooth but not continuous. Name one that is continuous but not smooth. 131. Which of the following statements are true regarding the graph of the cubic polynomial f 1x2 = x3 + bx2 + cx + d? (Give reasons for your conclusions.) (a) It intersects the y-axis in one and only one point. (b) It intersects the x-axis in at most three points. (c) It intersects the x-axis at least once. (d) For x very large, it behaves like the graph of y = x3. (e) It is symmetric with respect to the origin. (f) It passes through the origin. x (a) (b) (c) (d) (e) (f) Is the degree of the polynomial even or odd? Is the leading coefficient positive or negative? Is the function even, odd, or neither? Why is x2 necessarily a factor of the polynomial? What is the minimum degree of the polynomial? Formulate five different polynomials whose graphs could look like the one shown. Compare yours to those of other students. What similarities do you see? What differences? 133. Design a polynomial function with the following characteristics: degree 6; four distinct real zeros, one of multiplicity 3; y-intercept 3; behaves like y = - 5x6 for large values of x . Is this polynomial unique? Compare your polynomial with those of other students. What terms will be the same as everyone else’s? Add some more characteristics, such as symmetry or naming the real zeros. How does this modify the polynomial? Retain Your Knowledge Problems 134–137 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 134. Find the equation of the line that contains the point 12, - 32 and is perpendicular to the line 5x - 2y = 6. x - 3 . 135. Find the domain of the function h1x2 = x + 5 136. Use the quadratic formula to find the zeros of the function f 1x2 = 4x2 + 8x - 3. 137. Solve: 5x - 3 = 7. ‘Are You Prepared?’ Answers 1. 1 - 2, 02, 12, 02, 10, 92 5. True 6. (5, 0); 5 2. Yes; 3 3. Down; 4 4. Local maximum value 6.48 at x = - 0.67; local minimum value - 3 at x = 2 SECTION 5.2 Properties of Rational Functions 343 5.2 Properties of Rational Functions PREPARING FOR THIS SECTION Before getting started, review the following: r Graph of f 1x2 = r Rational Expressions (Chapter R, Section R.7, pp. 62–69) r Polynomial Division (Chapter R, Section R.4, pp. 44–47) 1 (Section 2.2, Example 12, p. 164) x r Graphing Techniques: Transformations (Section 3.5, pp. 247–256) Now Work the ‘Are You Prepared?’ problems on page 350. OBJECTIVES 1 Find the Domain of a Rational Function (p. 343) 2 Find the Vertical Asymptotes of a Rational Function (p. 346) 3 Find the Horizontal or Oblique Asymptote of a Rational Function (p. 348) Ratios of integers are called rational numbers. Similarly, ratios of polynomial functions are called rational functions. Examples of rational functions are R 1x2 = DEFINITION x2 - 4 x + x + 1 2 F1x2 = x3 x - 4 2 G1x2 = 3x2 x - 1 4 A rational function is a function of the form R 1x2 = p 1x2 q 1x2 where p and q are polynomial functions and q is not the zero polynomial. The domain of a rational function is the set of all real numbers except those for which the denominator q is 0. 1 Find the Domain of a Rational Function EXAMPL E 1 Finding the Domain of a Rational Function 2x2 - 4 is the set of all real numbers x except - 5; that x + 5 is, the domain is 5 x x ≠ - 56 . (a) The domain of R 1x2 = (b) The domain of R 1x2 = 1 1 = is the set of all real numbers x 1x + 22 1x - 22 x - 4 except - 2 and 2; that is, the domain is 5 x x ≠ - 2, x ≠ 26 . (c) The domain of R 1x2 = 2 x3 is the set of all real numbers. x2 + 1 x2 - 1 is the set of all real numbers x except 1; that is, x - 1 the domain is 5 x x ≠ 16 . (d) The domain of R 1x2 = r x2 - 1 Although reduces to x + 1, it is important to observe that the functions x - 1 R 1x2 = x2 - 1 x - 1 and f1x2 = x + 1 are not equal, since the domain of R is 5 x x ≠ 16 and the domain of f is the set of all real numbers. Now Work PROBLEM 17 344 CHAPTER 5 Polynomial and Rational Functions WARNING The domain of a rational function must be found before writing the function in lowest terms. ■ EX AMPLE 2 p 1x2 is a rational function, and if p and q have no common factors, q 1x2 then the rational function R is said to be in lowest terms. For a rational function p 1x2 R 1x2 = in lowest terms, the real zeros, if any, of the numerator in the domain q 1x2 of R are the x-intercepts of the graph of R and so will play a major role in the graph of R. The real zeros of the denominator of R [that is, the numbers x, if any, for which q 1x2 = 0], although not in the domain of R, also play a major role in the graph of R. 1 We have already discussed the properties of the rational function y = . x (Refer to Example 12, page 164). The next rational function that we take up is 1 H1x2 = 2 . x If R 1x2 = Graphing y = 1 x2 1 . x2 Analyze the graph of H1x2 = Solution 1 is the set of all real numbers x except 0. The graph has x2 no y-intercept, because x can never equal 0. The graph has no x-intercept because the equation H1x2 = 0 has no solution. Therefore, the graph of H will not cross or touch either of the coordinate axes. Because The domain of H1x2 = H1 - x2 = Table 7 x H(x) = 1 x2 1 2 4 1 100 10,000 1 10,000 100,000,000 1 1 2 100 10,000 H is an even function, so its graph is symmetric with respect to the y-axis. 1 Table 7 shows the behavior of H1x2 = 2 for selected positive numbers x. (We x use symmetry to obtain the graph of H when x 6 0.) From the first three rows of Table 7, we see that as the values of x approach (get closer to) 0, the values of H1x2 become larger and larger positive numbers, so H is unbounded in the positive direction. In calculus we use limit notation, lim H1x2 = q , which is read “the limit xS0 of H1x2 as x approaches zero equals infinity,” to mean that H1x2 S q as x S 0. Look at the last four rows of Table 7. As x S q , the values of H1x2 approach 0 (the end behavior of the graph). In calculus, this is expressed by writing lim H1x2 = 0. Figure 25 shows the graph. Notice the use of red dashed lines to x Sq convey the ideas discussed above. x0 y 5 ( 1–2 , 4) 1 4 1 10,000 1 100,000,000 (1, 1) 1– 4 Figure 25 H(x) = ( 1–2 , 4) (1, 1) (2, 1–4 ) (2, ) 3 x y0 y 0 3 EX AMPLE 3 1 1 = 2 = H1x2 2 1 - x2 x 1 x2 r Using Transformations to Graph a Rational Function Graph the rational function: R 1x2 = 1 + 1 1x - 22 2 345 SECTION 5.2 Properties of Rational Functions The domain of R is the set of all real numbers except x = 2. To graph R, start with 1 the graph of y = 2 . See Figure 26 for the steps. x Solution x⫽2 x⫽0 x⫽2 y y y 3 3 (3, 2) (⫺1, 1) 1 (1, 1) y ⫽ 0 ⫺2 3 y⫽0 x x 5 Replace x by x ⫺ 2; shift right 2 units (a) y ⫽ Figure 26 (1, 2) y⫽1 (3, 1) (1, 1) 1 x2 PROBLEMS x Add 1; shift up 1 unit (b) y ⫽ Now Work 5 1 (x – 2)2 35(a) (c) y ⫽ 1 ⫹1 (x ⫺ 2)2 35(b) AND r Asymptotes Let’s investigate the roles of the vertical line x = 2 and the horizontal line y = 1 in Figure 26(c). 1 First, we look at the end behavior of R 1x2 = + 1. Table 8(a) shows 1x - 22 2 the values of R at x = 10, 100, 1000, and 10,000. Note that as x becomes unbounded in the positive direction, the values of R approach 1, so lim R 1x2 = 1. From Table 8(b) x Sq we see that as x becomes unbounded in the negative direction, the values of R also approach 1, so lim R 1x2 = 1. xS -q Even though x = 2 is not in the domain of R, the behavior of the graph of R near x = 2 is important. Table 8(c) shows the values of R at x = 1.5, 1.9, 1.99, 1.999, and 1.9999. We see that as x approaches 2 for x 6 2, denoted x S 2 - , the values of R are increasing without bound, so lim- R 1x2 = q . From Table 8(d), we see that xS2 as x approaches 2 for x 7 2, denoted x S 2 + , the values of R are also increasing without bound, so lim+ R 1x2 = q . Table 8 xS2 x R(x) x R(x) x R(x) x R(x) 10 1.0156 - 10 1.0069 1.5 5 2.5 5 100 1.0001 - 100 1.0001 1.9 101 2.1 101 1000 1.000001 - 1000 1.000001 1.99 10,001 2.01 10,001 10,000 1.00000001 - 10,000 1.00000001 1.999 1,000,001 2.001 1,000,001 1.9999 100,000,001 2.0001 100,000,001 (a) (b) (c) (d) The vertical line x = 2 and the horizontal line y = 1 are called asymptotes of the graph of R. DEFINITION Let R denote a function. If, as x S - q or as x S q , the values of R 1x2 approach some fixed number L, then the line y = L is a horizontal asymptote of the graph of R. [Refer to Figures 27(a) and (b) on page 346.] If, as x approaches some number c, the values R 1x2 S q [that is, R 1x2 S - q or R 1x2 S q ], then the line x = c is a vertical asymptote of the graph of R. [Refer to Figures 27(c) and (d).] 346 CHAPTER 5 Polynomial and Rational Functions x5c y y y y x5c y 5 R(x ) y5L y5L x D y 5 R(x ) End behavior: As x → `, the values of R(x ) approach L [ xlim R(x) 5 L]. →` That is, the points on the graph of R are getting closer to the line y 5 L; y 5 L is a horizontal asymptote. E End behavior: As x → 2`, the values of R(x) approach L lim R (x) 5 L]. That is, the [ x→ 2` points on the graph of R are getting closer to the line y 5 L; y 5 L is a horizontal asymptote. x x x F As x approaches c, the values of ⏐R (x)⏐→ ` [ xlim → c2 R(x) 5 `; lim x → c 1R(x) 5 `]. That is, the points on the graph of R are getting closer to the line x 5 c ; x 5 c is a vertical asymptote. G As x approaches c, the values of ⏐R(x)⏐→ ` [ xlim → c2 R(x) 5 2`; lim R(x) 5 `]. That is, x → c1 the points on the graph of R are getting closer to the line x 5 c; x 5 c is a vertical asymptote. Figure 27 y x Figure 28 Oblique asymptote A horizontal asymptote, when it occurs, describes the end behavior of the graph as x S q or as x S - q . The graph of a function may intersect a horizontal asymptote. A vertical asymptote, when it occurs, describes the behavior of the graph when x is close to some number c. The graph of a rational function will never intersect a vertical asymptote. There is a third possibility. If, as x S - q or as x S q , the value of a rational function R 1x2 approaches a linear expression ax + b, a ≠ 0, then the line y = ax + b, a ≠ 0, is an oblique (or slant) asymptote of R. Figure 28 shows an oblique asymptote. An oblique asymptote, when it occurs, describes the end behavior of the graph. The graph of a function may intersect an oblique asymptote. Now Work PROBLEMS 27 AND 35(c) 2 Find the Vertical Asymptotes of a Rational Function p 1x2 , in lowest terms, are q 1x2 located at the real zeros of the denominator q 1x2. Suppose that r is a real zero of q, so x - r is a factor of q. As x approaches r, symbolized as x S r, the values of x - r approach 0, causing the ratio to become unbounded; that is, R 1x2 S q . Based on the definition, we conclude that the line x = r is a vertical asymptote. The vertical asymptotes of a rational function R 1x2 = THEOREM WARNING If a rational function is not in lowest terms, an application of this theorem may result in an incorrect listing of vertical asymptotes. ■ EX AMPLE 4 Locating Vertical Asymptotes p 1x2 , in lowest terms, will have a vertical asymptote A rational function R 1x2 = q 1x2 x = r if r is a real zero of the denominator q. That is, if x - r is a factor of the p 1x2 denominator q of a rational function R 1x2 = , in lowest terms, R will q 1x2 have the vertical asymptote x = r. Finding Vertical Asymptotes Find the vertical asymptotes, if any, of the graph of each rational function. x + 3 x - 1 x2 (c) H1x2 = 2 x + 1 (a) F1x2 = (b) R 1x2 = x x - 4 x2 - 9 (d) G1x2 = 2 x + 4x - 21 2 SECTION 5.2 Properties of Rational Functions Solution WARNING In Example 4(a), the vertical asymptote is x = 1. Do not say that the vertical asymptote is 1. ■ 347 (a) F is in lowest terms, and the only zero of the denominator is 1. The line x = 1 is the vertical asymptote of the graph of F. (b) R is in lowest terms, and the zeros of the denominator x2 - 4 are - 2 and 2. The lines x = - 2 and x = 2 are the vertical asymptotes of the graph of R. (c) H is in lowest terms, and the denominator has no real zeros because the equation x2 + 1 = 0 has no real solutions. The graph of H has no vertical asymptotes. (d) Factor the numerator and denominator of G1x2 to determine whether it is in lowest terms. 1x + 32 1x - 32 x2 - 9 x + 3 = = x ≠ 3 1x + 72 1x - 32 x + 7 x + 4x - 21 The only zero of the denominator of G1x2 in lowest terms is - 7. The line x = - 7 is the only vertical asymptote of the graph of G. G1x2 = 2 r As Example 4 points out, rational functions can have no vertical asymptotes, one vertical asymptote, or more than one vertical asymptote. Multiplicity and Vertical Asymptotes Recall from Figure 15 in Section 5.1 that the end behavior of a polynomial function is always one of four types. For polynomials of odd degree, the ends of the graph go in opposite directions (one up and one down), whereas for polynomials of even degree, the ends go in the same direction (both up or both down). For a rational function in lowest terms, the multiplicities of the zeros in the denominator can be used in a similar fashion to determine the behavior of the graph around each vertical asymptote. Consider the following four functions, each with a single vertical asymptote, x = 2. 1 1 1 1 R2 1x2 = R3 1x2 = R4 1x2 = R1 1x2 = x - 2 x - 2 1x - 22 2 1x - 22 2 Figure 29 shows the graphs of each function. The graphs of R1 and R2 are 1 transformations of the graph of y = , and the graphs of R3 and R4 are x 1 transformations of the graph of y = 2 . x Based on Figure 29, we can make the following conclusions: r If the multiplicity of the zero that gives rise to a vertical asymptote is odd, the graph approaches q on one side of the vertical asymptote and approaches - q on the other side. r If the multiplicity of the zero that gives rise to the vertical asymptote is even, the graph approaches either q or - q on both sides of the vertical asymptote. These results are true in general and will be helpful when graphing rational functions in the next section. y 6 y 6 R1(x) −1 5 −6 x2 (a) Odd multiplicity lim- R1(x) = - q xS2 lim+ R1(x) = q xS2 Figure 29 y 6 x2 y 6 R3(x) x2 y0 y0 y0 y0 x x x x −1 5 −6 R2(x) (b) Odd multiplicity lim- R2(x) = q xS2 lim+ R2(x) = - q xS2 −1 5 −6 x2 (c) Even multiplicity lim- R3(x) = q xS2 lim+ R3(x) = q xS2 −1 5 −6 R4(x) (d) Even multiplicity lim- R4(x) = - q xS2 lim R4(x) = - q xS2 + 348 CHAPTER 5 Polynomial and Rational Functions 3 Find the Horizontal or Oblique Asymptote of a Rational Function To find horizontal or oblique asymptotes, we need to know how the value of the function behaves as x S - q or as x S q . That is, we need to determine the end behavior of the function. This can be done by examining the degrees of the numerator and denominator, and the respective power functions that each resembles. For example, consider the rational function R 1x2 = 3x - 2 5x - 7x + 1 2 The degree of the numerator, 1, is less than the degree of the denominator, 2. When x is very large, the numerator of R can be approximated by the power function y = 3x, and the denominator can be approximated by the power function y = 5x2. This means R 1x2 = 3x - 2 3x 3 S0 ≈ = 2 5x c 5x - 7x + 1 c 5x 2 For |x| very large As x S - q or x S q which shows that the line y = 0 is a horizontal asymptote. This result is true for all rational functions that are proper (that is, the degree of the numerator is less than the degree of the denominator). If a rational function is improper (that is, if the degree of the numerator is greater than or equal to the degree of the denominator), there could be a horizontal asymptote, an oblique asymptote, or neither. The following summary details how to find horizontal or oblique asymptotes. Finding a Horizontal or Oblique Asymptote of a Rational Function Consider the rational function R 1x2 = p 1x2 anxn + an - 1xn - 1 + g + a1x + a0 = q 1x2 bmxm + bm - 1xm - 1 + g + b1x + b0 in which the degree of the numerator is n and the degree of the denominator is m. 1. If n 6 m (the degree of the numerator is less than the degree of the denominator), the line y = 0 is a horizontal asymptote. 2. If n = m (the degree of the numerator equals the degree of the an denominator), the line y = is a horizontal asymptote. (That is, the bm horizontal asymptote equals the ratio of the leading coefficients.) 3. If n = m + 1 (the degree of the numerator is one more than the degree of the denominator), the line y = ax + b is an oblique asymptote, which is the quotient found using long division. 4. If n Ú m + 2 (the degree of the numerator is two or more greater than the degree of the denominator), there are no horizontal or oblique asymptotes. The end behavior of the graph will resemble the power function an n - m y = x . bm Note: A rational function will never have both a horizontal asymptote and an oblique asymptote. A rational function may have neither a horizontal nor an oblique asymptote. We illustrate each of the possibilities in Examples 5 through 8. SECTION 5.2 Properties of Rational Functions EXAMPL E 5 349 Finding a Horizontal Asymptote Find the horizontal asymptote, if one exists, of the graph of R 1x2 = Solution 4x3 - 5x + 2 7x5 + 2x4 - 3x Since the degree of the numerator, 3, is less than the degree of the denominator, 5, the rational function R is proper. The line y = 0 is a horizontal asymptote of the graph of R. r EXAMPL E 6 Finding a Horizontal or Oblique Asymptote Find the horizontal or oblique asymptote, if one exists, of the graph of H1x2 = Solution 3x4 - x2 x3 - x2 + 1 Since the degree of the numerator, 4, is exactly one greater than the degree of the denominator, 3, the rational function H has an oblique asymptote. Find the asymptote by using long division. 3x + 3 - x2 x3 - x2 + 1 ) 3x4 3x4 - 3x3 + 3x 3x3 - x2 - 3x 3x3 - 3x2 + 3 2x2 - 3x - 3 As a result, H1x2 = 3x4 - x2 2x2 - 3x - 3 = 3x + 3 + x3 - x2 + 1 x3 - x2 + 1 As x S - q or as x S q , 2x2 - 3x - 3 2x2 2 ≈ = S0 3 2 3 x x - x + 1 x As x S - q or as x S q , we have H1x2 S 3x + 3. The graph of the rational function H has an oblique asymptote y = 3x + 3. Put another way, as x S { q , the graph of H will behave like the graph of y = 3x + 3. r EXAMPL E 7 Finding a Horizontal or Oblique Asymptote Find the horizontal or oblique asymptote, if one exists, of the graph of R 1x2 = Solution 8x2 - x + 2 4x2 - 1 Since the degree of the numerator, 2, equals the degree of the denominator, 2, the rational function R has a horizontal asymptote equal to the ratio of the leading coefficients. y = an 8 = = 2 bm 4 To see why the horizontal asymptote equals the ratio of the leading coefficients, investigate the behavior of R as x S - q or as x S q . When x is very large, the numerator of R can be approximated by the power function y = 8x2, and the 350 CHAPTER 5 Polynomial and Rational Functions denominator can be approximated by the power function y = 4x2. This means that as x S - q or as x S q , R 1x2 = 8 8x2 8x2 - x + 2 = = 2 ≈ 2 2 4 4x - 1 4x The graph of the rational function R has a horizontal asymptote y = 2. The graph of R will behave like y = 2 as x S { q . r EX AMPLE 8 Finding a Horizontal or Oblique Asymptote Find the horizontal or oblique asymptote, if one exists, of the graph of G1x2 = Solution 2x5 - x3 + 2 x3 - 1 Since the degree of the numerator, 5, is greater than the degree of the denominator, 3, by more than one, the rational function G has no horizontal or oblique asymptote. The end behavior of the graph will resemble the power function y = 2x5-3 = 2x2. To see why this is the case, investigate the behavior of G as x S - q or as x S q . When x is very large, the numerator of G can be approximated by the power function y = 2x5, and the denominator can be approximated by the power function y = x3. This means as x S - q or as x S q , G1x2 = 2x5 2x5 - x3 + 2 ≈ = 2x5 - 3 = 2x2 x3 - 1 x3 Since this is not linear, the graph of G has no horizontal or oblique asymptote. The graph of G will behave like y = 2x2 as x S { q . r Now Work PROBLEMS 45, 47, AND 49 5.2 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. True or False The quotient of two polynomial expressions is a rational expression. (pp. 62–69) 3. Graph y = 2. What are the quotient and remainder when 3x4 - x2 is divided by x3 - x2 + 1. (pp. 44–47) 4. Graph y = 21x + 12 2 - 3 using transformations. 1 . (p. 164) x (pp. 247–256) Concepts and Vocabulary 5. True or False The domain of every rational function is the set of all real numbers. 6. If, as x S - q or as x S q , the values of R 1x2 approach some fixed number L, then the line y = L is a of the graph of R. 7. If, as x approaches some number c, the values of R 1x2 S q , then the line x = c is a of the graph of R. 8. For a rational function R, if the degree of the numerator is less than the degree of the denominator, then R is . 9. True or False The graph of a rational function may intersect a horizontal asymptote. 10. True or False The graph of a rational function may intersect a vertical asymptote. 11. If a rational function is proper, then asymptote. is a horizontal 12. True or False If the degree of the numerator of a rational function equals the degree of the denominator, then the ratio of the leading coefficients gives rise to the horizontal asymptote. p1x2 13. If R 1x2 = is a rational function and if p and q have no q1x2 common factors, then R is . (a) improper (c) undefined (b) proper (d) in lowest terms 14. Which type of asymptote, when it occurs, describes the behavior of a graph when x is close to some number? (a) vertical (b) horizontal (c) oblique (d) all of these 351 SECTION 5.2 Properties of Rational Functions Skill Building In Problems 15–26, find the domain of each rational function. 15. R 1x2 = 4x x - 3 16. R 1x2 = 18. G1x2 = 6 1x + 32 14 - x2 19. F 1x2 = 21. R 1x2 = x x - 8 22. R 1x2 = x x - 1 23. H1x2 = 24. G1x2 = x - 3 x4 + 1 25. R 1x2 = 31x2 - x - 62 26. F 1x2 = 3 In Problems 27–32, use the graph shown to find (a) The domain and range of each function (d) Vertical asymptotes, if any 27. 5x2 3 + x 3x1x - 12 20. Q1x2 = 2 2x - 5x - 3 4 2 41x - 92 (b) The intercepts, if any (e) Oblique asymptotes, if any 28. y 17. H1x2 = - x11 - x2 3x2 + 5x - 2 3x2 + x x2 + 4 - 21x2 - 42 31x2 + 4x + 42 (c) Horizontal asymptotes, if any 29. y 4 - 4x2 1x - 22 1x + 42 y 3 3 (0, 2) 4 x –4 3 (1, 2) (1, 0) (1, 0) 3 x 3 –4 3 x 3 3 30. 31. 32. y y y 3 (1, 2) 3 3 (1, 1) 3 3 3 x 3 3 3 x 3 (1, 2) 3 3 In Problems 33–44, (a) graph the rational function using transformations, (b) use the final graph to find the domain and range, and (c) use the final graph to list any vertical, horizontal, or oblique asymptotes. 33. F 1x2 = 2 + 37. H1x2 = 1 x -2 x + 1 41. G1x2 = 1 + 2 1x - 32 2 34. Q1x2 = 3 + 38. G1x2 = 1 x2 2 1x + 22 2 42. F 1x2 = 2 - 1 x + 1 35. R 1x2 = 1 1x - 12 2 36. R 1x2 = 3 x 39. R 1x2 = -1 x + 4x + 4 40. R 1x2 = 1 + 1 x - 1 43. R 1x2 = x2 - 4 x2 44. R 1x2 = x - 4 x 2 x 352 CHAPTER 5 Polynomial and Rational Functions In Problems 45–56, find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. 45. R 1x2 = 3x x + 4 46. R 1x2 = 3x + 5 x - 6 47. H1x2 = x3 - 8 x - 5x + 6 48. G1x2 = x3 + 1 x - 5x - 14 49. T 1x2 = x3 x - 1 50. P 1x2 = 4x2 x - 1 51. Q1x2 = 2x2 - 5x - 12 3x2 - 11x - 4 52. F 1x2 = x2 + 6x + 5 2x2 + 7x + 5 53. R 1x2 = 6x2 + 7x - 5 3x + 5 54. R 1x2 = 8x2 + 26x - 7 4x - 1 55. G1x2 = x4 - 1 x2 - x 56. F 1x2 = x4 - 16 x2 - 2x 4 3 2 2 Applications and Extensions 57. Gravity In physics, it is established that the acceleration due to gravity, g (in meters/sec2), at a height h meters above sea level is given by 60. Newton’s Method In calculus you will learn that if p1x2 = anxn + an-1xn-1 + g + a1x + a0 is a polynomial function, then the derivative of p1x2 is g1h2 = 3.99 * 1014 16.374 * 10 + h2 6 2 where 6.374 * 106 is the radius of Earth in meters. (a) What is the acceleration due to gravity at sea level? (b) The Willis Tower in Chicago, Illinois, is 443 meters tall. What is the acceleration due to gravity at the top of the Willis Tower? (c) The peak of Mount Everest is 8848 meters above sea level. What is the acceleration due to gravity on the peak of Mount Everest? (d) Find the horizontal asymptote of g1h2 . (e) Solve g1h2 = 0. How do you interpret your answer? 58. Population Model A rare species of insect was discovered in the Amazon Rain Forest. To protect the species, environmentalists declared the insect endangered and transplanted the insect into a protected area.The population P of the insect t months after being transplanted is P 1t2 = 5011 + 0.5t2 2 + 0.01t (a) How many insects were discovered? In other words, what was the population when t = 0? (b) What will the population be after 5 years? (c) Determine the horizontal asymptote of P 1t2 . What is the largest population that the protected area can sustain? 59. Resistance in Parallel Circuits From Ohm’s Law for circuits, it follows that the total resistance Rtot of two components hooked in parallel is given by the equation Rtot = R1R2 R1 + R2 where R1 and R2 are the individual resistances. (a) Let R1 = 10 ohms, and graph Rtot as a function of R2. (b) Find and interpret any asymptotes of the graph obtained in part (a). (c) If R2 = 21R1, what value of R1 will yield an Rtot of 17 ohms? p′ 1x2 = nanxn-1 + 1n - 12an-1xn-2 + g + 2a2x + a1 Newton’s Method is an efficient method for approximating the x-intercepts (or real zeros) of a function, such as p1x2 . The following steps outline Newton’s Method. STEP 1: Select an initial value x0 that is somewhat close to the x-intercept being sought. STEP 2: Find values for x using the relation xn + 1 = xn - p1xn 2 p′ 1xn 2 n = 1, 2, c until you get two consecutive values xn and xn + 1 that agree to whatever decimal place accuracy you desire. STEP 3: The approximate zero will be xn + 1. Consider the polynomial p1x2 = x3 - 7x - 40. (a) Evaluate p152 and p1 - 32 . (b) What might we conclude about a zero of p? Explain. (c) Use Newton’s Method to approximate an x-intercept, r, - 3 6 r 6 5, of p1x2 to four decimal places. (d) Use a graphing utility to graph p1x2 and verify your answer in part (c). (e) Using a graphing utility, evaluate p1r2 to verify your result. 61. Exploration The standard form of the rational mx + b function R 1x2 = , where c ≠ 0, cx + d 1 is R 1x2 = a¢ ≤ + k. To write a rational function x - h in standard form requires long division. 2x + 3 (a) Write the rational function R 1x2 = in x - 1 standard form by writing R in the form Quotient + remainder divisor (b) Graph R using transformations. (c) Determine the vertical asymptote and the horizontal asymptote of R. 62. Exploration Repeat Problem 61 - 6x + 16 function R 1x2 = . 2x - 7 for the rational SECTION 5.3 The Graph of a Rational Function 353 Explaining Concepts: Discussion and Writing 63. If the graph of a rational function R has the vertical asymptote x = 4, the factor x - 4 must be present in the denominator of R. Explain why. 64. If the graph of a rational function R has the horizontal asymptote y = 2, the degree of the numerator of R equals the degree of the denominator of R. Explain why. 65. The graph of a rational function cannot have both a horizontal and an oblique asymptote? Explain why. 66. Make up a rational function that has y = 2x + 1 as an oblique asymptote. Explain the methodology that you used. Retain Your Knowledge Problems 67–70 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x 2 68. Solve: 13x - 72 + 1 = - 2 67. Find the equation of a vertical line passing through the 5 4 point 15, - 32 . 69. Determine whether the graph of the equation 2x3 - xy2 = 4 is symmetric with respect to the x-axis, the y-axis, the origin, or none of these. 70. What are the points of intersection of the graphs of the functions f 1x2 = - 3x + 2 and g1x2 = x2 - 2x - 4? ‘Are You Prepared?’ Answers 1. True 2. Quotient: 3x + 3; remainder: 2x2 - 3x - 3 y 3. y 3 4. 2 (1, 1) 2 3 2 x 3 x (0,1) (1, 1) 2 (1,3) 3 5.3 The Graph of a Rational Function PREPARING FOR THIS SECTION Before getting started, review the following: r Intercepts Section 2.2, pp. 159–160 Now Work the ‘Are You Prepared?’ problem on page 365. OBJECTIVES 1 Analyze the Graph of a Rational Function (p. 353) 2 Solve Applied Problems Involving Rational Functions (p. 364) 1 Analyze the Graph of a Rational Function We commented earlier that calculus provides the tools required to graph a polynomial function accurately. The same holds true for rational functions. However, we can gather together quite a bit of information about their graphs to get an idea of the general shape and position of the graph. EXAMPL E 1 How to Analyze the Graph of a Rational Function Analyze the graph of the rational function: R 1x2 = Step-by-Step Solution Step 1: Factor the numerator and denominator of R. Find the domain of the rational function. R 1x2 = x - 1 x2 - 4 x - 1 x - 1 = 2 1x + 22 1x - 22 x - 4 The domain of R is {x x ≠ - 2, x ≠ 2}. 354 CHAPTER 5 Polynomial and Rational Functions Step 2: Write R in lowest terms. Step 3: Find and plot the intercepts of the graph. Use multiplicity to determine the behavior of the graph of R at each x-intercept. Because there are no common factors between the numerator and denominator, R is in lowest terms. 1 1 . Plot the point ¢ 0, ≤. The 4 4 x-intercepts are found by determining the real zeros of the numerator of R written in lowest terms. By solving x - 1 = 0, we find that the only real zero of the numerator is 1, so the only x-intercept of the graph of R is 1. Plot the point (1, 0). The multiplicity of 1 is odd, so the graph will cross the x-axis at x = 1. Since 0 is in the domain of R, the y-intercept is R(0) = Step 4: Find the vertical asymptotes. Graph each vertical asymptote using a dashed line. Determine the behavior of the graph on either side of each vertical asymptote. The vertical asymptotes are the zeros of the denominator with the rational function in lowest terms. With R written in lowest terms, we find that the graph of R has two vertical asymptotes: the lines x = - 2 and x = 2. See Figure 30(a). The multiplicities of the zeros that give rise to the vertical asymptotes are both odd. Therefore, the graph will approach q on one side of each vertical asymptote, and will approach - q on the other side. Step 5: Find the horizontal or oblique asymptote, if one exists. Find points, if any, at which the graph of R intersects this asymptote. Graph the asymptote using a dashed line. Plot any points at which the graph of R intersects the asymptote. Because the degree of the numerator is less than the degree of the denominator, R is proper and the line y = 0 (the x-axis) is a horizontal asymptote of the graph. To determine whether the graph of R intersects the horizontal asymptote, solve the equation R 1x2 = 0: x - 1 = 0 x2 - 4 x - 1 = 0 x = 1 The only solution is x = 1, so the graph of R intersects the horizontal asymptote at 11, 02. Step 6: Use the zeros of the numerator and denominator of R to divide the x-axis into intervals. Determine where the graph of R is above or below the x-axis by choosing a number in each interval and evaluating R there. Plot the points found. The zero of the numerator, 1, and the zeros of the denominator, - 2 and 2, divide the x-axis into four intervals: 1 - q , - 22 1 - 2, 12 11, 22 12, q 2 Now construct Table 9. Table 9 –2 1 2 x Interval ( - q , - 2) ( - 2, 1) (1, 2) (2, q ) Number chosen -3 -1 3 2 3 Value of R R( - 3) = - 0.8 R( - 1) = 2 3 Ra b = 2 7 R(3) = 0.4 Location of graph Below x-axis Above x-axis Below x-axis Above x-axis Point on graph ( - 3, - 0.8) 2 a - 1, b 3 3 2 a ,- b 2 7 (3, 0.4) 2 3 Figure 30(a) shows the asymptotes, the points from Table 9, the y-intercept, and the x-intercept. Step 7: Use the results obtained in Steps 1 through 6 to graph R. r The graph crosses the x-axis at x = 1, changing from being above the x-axis for x 6 1 to below it for x 7 1. Indicate this on the graph. See Figure 30(b). r Since y = 0 (the x-axis) is a horizontal asymptote and the graph lies below the x-axis for x 6 - 2, we can sketch a portion of the graph by placing a small arrow to the far left and under the x-axis. SECTION 5.3 The Graph of a Rational Function 355 r Since the line x = - 2 is a vertical asymptote and the graph lies below the x-axis for x 6 - 2, we place an arrow well below the x-axis and approaching the line x = - 2 from the left 1 lim -R 1x2 = - q 2 . x S -2 r Since the graph approaches - q on one side of x = - 2, and - 2 is a zero of odd multiplicity, the graph will approach q on the other side of x = - 2. That is, lim +R 1x2 = q . Similar analysis leads to lim-R(x) = - q and lim+R(x) = q . x S -2 Finally, lim R(x) = 0 and lim R(x) = 0. xS -q xS 2 xS 2 xS q Figure 30(b) illustrates these conclusions and Figure 30(c) shows the graph of R. x = −2 x=2 y x = −2 3 (1, 0) (−3, −0.8) x = −2 (0, 1–4) ( 3–2 ,−2–7) 3 (1, 0) xy = 0 −3 (−3, −0.8) (0, 1–4) ( 3–2 ,−2–7) (3, 0.4) xy = 0 3 (−1, 2–3) −3 x=2 (−3, −0.8) (1, 0) (0, 1–4) ( 3– ,−2–) 7 2 (3, 0.4) 3 xy=0 −3 −3 (a) y 3 (−1, 2–3) (3, 0.4) −3 Figure 30 x=2 3 (−1, 2–3) −3 y (b) r (c) Exploration Graph the rational function: R(x) = x - 1 x2 - 4 Result The analysis just completed in Example 1 helps us to set the viewing rectangle to obtain a x - 1 complete graph. Figure 31(a) shows the graph of R(x) = 2 in connected mode, and Figure 31(b) x - 4 shows it in dot mode. Notice in Figure 31(a) that the graph has vertical lines at x = - 2 and x = 2. This is due to the fact that when a graphing utility is in connected mode, some will connect the dots between consecutive pixels, and vertical lines may occur. We know that the graph of R does not cross the lines x = - 2 and x = 2, since R is not defined at x = - 2 or x = 2. So, when graphing rational functions, use dot mode if extraneous vertical lines are present in connected mode. Newer graphing utilities may not have extraneous vertical lines in connected mode. See Figure 31(c). 4 4 24 4 24 4 4 24 4 24 Figure 31 24 Connected mode with extraneous vertical lines (a) 24 Dot mode (b) Connected mode without extraneous vertical lines (c) 356 CHAPTER 5 Polynomial and Rational Functions SUMMARY Analyzing the Graph of a Rational Function R STEP 1: Factor the numerator and denominator of R. Find the domain of the rational function. STEP 2: Write R in lowest terms. STEP 3: Find and plot the intercepts of the graph. Use multiplicity to determine the behavior of the graph of R at each x-intercept. STEP 4: Find the vertical asymptotes. Graph each vertical asymptote using a dashed line. Determine the behavior of the graph of R on either side of each vertical asymptote. STEP 5: Find the horizontal or oblique asymptote, if one exists. Find points, if any, at which the graph of R intersects this asymptote. Graph the asymptote using a dashed line. Plot any points at which the graph of R intersects the asymptote. STEP 6: Use the zeros of the numerator and denominator of R to divide the x-axis into intervals. Determine where the graph of R is above or below the x-axis by choosing a number in each interval and evaluating R there. Plot the points found. STEP 7: Use the results obtained in Steps 1 through 6 to graph R. Now Work EX AMPLE 2 PROBLEM 7 Analyzing the Graph of a Rational Function Analyze the graph of the rational function: R 1x2 = Solution STEP 4: STEP 5: x2 - 1 x2 1 1 = = x x x x x 1 Since S 0 as x S q , y = x is the x ■ oblique asymptote. 1x + 12 1x - 12 . The domain of R is 5 x x ≠ 06 . x R is in lowest terms. Because x cannot equal 0, there is no y-intercept. The graph has two x-intercepts, - 1 and 1, each with odd multiplicity. Plot the points 1 - 1, 02 and 11, 02 . The graph will cross the x-axis at both points. The real zero of the denominator with R in lowest terms is 0, so the graph of R has the line x = 0 (the y-axis) as a vertical asymptote. Graph x = 0 using a dashed line. The multiplicity of 0 is odd, so the graph will approach q on one side of the asymptote x = 0, and - q on the other side. Since the degree of the numerator, 2, is one greater than the degree of the denominator, 1, the rational function will have an oblique asymptote. To find the oblique asymptote, use long division. STEP 1: R 1x2 = STEP 2: STEP 3: NOTE Because the denominator of the rational function is a monomial, we can also find the oblique asymptote as follows: x2 - 1 x x x ) x2 - 1 x2 -1 The quotient is x, so the line y = x is an oblique asymptote of the graph. Graph y = x using a dashed line. To determine whether the graph of R intersects the asymptote y = x, solve the equation R 1x2 = x. R 1x2 = x2 - 1 = x x x2 - 1 = x2 - 1 = 0 Impossible x2 - 1 The equation = x has no solution, so the graph of R does not intersect x the line y = x. 357 SECTION 5.3 The Graph of a Rational Function STEP 6: The zeros of the numerator are - 1 and 1; the zero of the denominator is 0. Use these values to divide the x-axis into four intervals: 1 - q , - 12 1 - 1, 02 10, 12 11, q 2 Now construct Table 10. Plot the points from Table 10. You should now have Figure 32(a). –1 Table 10 ( - 1, 0) (0, 1) Number chosen -2 1 2 1 2 2 Value of R R( - 2) = - 3 1 Ra - b = 2 2 1 3 Ra b = 2 2 R(2) = Location of graph Below x-axis Above x-axis Below x-axis Above x-axis Point on graph 3 a - 2, - b 2 1 3 a- , b 2 2 1 3 a ,- b 2 2 3 a2, b 2 x=0 y y y=x 3 (− 1–2 , 3–2 ) (2, ) 3 – 2 (−1, 0) (1, 0) ) ( 1–2 , −3–2) −3 (a) 3 x (− 1–2 , 3–2 ) (2, ) 3 – 2 (−2, (2, 3–2 ) (−1, 0) −3 −3–2 y=x 3 (−1, 0) −3 3 2 x=0 y y=x (− 1–2 , 3–2 ) Figure 32 3 2 Since the graph of R is above the x-axis for - 1 6 x 6 0, the graph of R will approach the vertical asymptote x = 0 at the top to the left of x = 0 [ lim-R 1x2 = q ]; since the graph of R approaches q on one side of the xS 0 asymptote and - q on the other, the graph of R will approach the vertical asymptote x = 0 at the bottom to the right of x = 0 [ lim+ R(x) = - q ]. xS 0 See Figure 32(b). The complete graph is given in Figure 32(c). x=0 3 x (1, q ) STEP 7: The graph crosses the x-axis at x = −1 and x = 1, changing from being below the x-axis to being above it in both cases. Since the graph of R is below the x-axis for x 6 - 1 and is above the x-axis for x 7 1, and since the graph of R does not intersect the oblique asymptote y = x, the graph of R will approach the line y = x as shown in Figure 32(b). NOTE Notice that R in Example 2 is an odd function. Do you see the symmetry about the origin in the graph of R in Figure 32(c)? ■ (−2, 1 ( - q , - 1) Interval −3–2 0 (1, 0) ) (1–2 , −3–2) −3 (b) 3 x −3 (−2, −3–2 (1, 0) ) 3 x ( 1–2 , −3–2 ) −3 (c) r 358 CHAPTER 5 Polynomial and Rational Functions Seeing the Concept x2 - 1 and compare what you see with Figure 32(c). Could you have predicted from the x graph that y = x is an oblique asymptote? Graph y = x and ZOOM-OUT. What do you observe? Graph R(x) = Now Work PROBLEM 15 Analyzing the Graph of a Rational Function EX AMPLE 3 Analyze the graph of the rational function: R(x) = Solution x0 y 6 y x2 STEP 1: R is completely factored. The domain of R is 5 x x ≠ 06 . STEP 2: R is in lowest terms. STEP 3: There is no y-intercept. Since x4 + 1 = 0 has no real solutions, there are no x-intercepts. STEP 4: R is in lowest terms, so x = 0 (the y-axis) is a vertical asymptote of R. Graph the line x = 0 using dashes. The multiplicity of 0 is even, so the graph will approach either q or - q on both sides of the asymptote. STEP 5: Since the degree of the numerator, 4, is two more than the degree of the denominator, 2, the rational function will not have a horizontal or oblique asymptote. Find the end behavior of R. As x S q , R 1x2 = (1, 2) (1, 2) 3 3 x (a) x4 + 1 x2 x4 + 1 x4 ≈ = x2 x2 x2 The graph of R will approach the graph of y = x2 as x S - q and as x S q . The graph of R does not intersect y = x2. Do you know why? Graph y = x2 using dashes. STEP 6: The numerator has no real zeros, and the denominator has one real zero at 0. Divide the x-axis into the two intervals 1 - q , 02 x0 10, q 2 and construct Table 11. y 6 y x2 0 Table 11 (1, 2) (1, 2) 3 3 x x Interval ( - q , 0) (0, q ) Number chosen -1 1 Value of R R( - 1) = 2 R(1) = 2 Location of graph Above x-axis Above x-axis Point on graph ( - 1, 2) (1, 2) (b) Figure 33 NOTE Notice that R in Example 3 is an even function. Do you see the symmetry about the y-axis in the graph of R? ■ STEP 7: Since the graph of R is above the x-axis and does not intersect y = x2, place arrows above y = x2 as shown in Figure 33(a). Also, since the graph of R is above the x-axis, and the multiplicity of the zero that gives rise to the vertical asymptote, x = 0, is even, it will approach the vertical asymptote x = 0 at the top to the left of x = 0 and at the top to the right of x = 0. See Figure 33(a). Figure 33(b) shows the complete graph. r SECTION 5.3 The Graph of a Rational Function 359 Seeing the Concept x4 + 1 and compare what you see with Figure 33(b). Use MINIMUM to find the two turning x2 2 points. Enter Y2 = x and ZOOM-OUT. What do you see? Graph R(x) = Now Work EXAMPL E 4 PROBLEM 13 Analyzing the Graph of a Rational Function Analyze the graph of the rational function: R 1x2 = Solution 3x2 - 3x x + x - 12 2 STEP 1: Factor R to get R 1x2 = STEP 2: STEP 3: STEP 4: STEP 5: 3x1x - 12 1x + 42 1x - 32 The domain of R is 5 x x ≠ - 4, x ≠ 36 . R is in lowest terms. The y-intercept is R 102 = 0. Plot the point 10, 02 . Since the real solutions of the equation 3x1x - 12 = 0 are x = 0 and x = 1, the graph has two x-intercepts, 0 and 1, each with odd multiplicity. Plot the points 10, 02 and 11, 02; the graph will cross the x-axis at both points. R is in lowest terms. The real solutions of the equation 1x + 42 1x - 32 = 0 are x = - 4 and x = 3, so the graph of R has two vertical asymptotes, the lines x = - 4 and x = 3. Graph these lines using dashes. The multiplicities that give rise to the vertical asymptotes are both odd, so the graph will approach q on one side of each vertical asymptote and - q on the other side. Since the degree of the numerator equals the degree of the denominator, the graph has a horizontal asymptote. To find it, form the quotient of the leading coefficient of the numerator, 3, and the leading coefficient of the denominator, 1. The graph of R has the horizontal asymptote y = 3. To find out whether the graph of R intersects the asymptote, solve the equation R 1x2 = 3. R 1x2 = 3x2 - 3x = 3 x + x - 12 3x2 - 3x = 3x2 + 3x - 36 - 6x = - 36 x = 6 2 The graph intersects the line y = 3 at x = 6, and 16, 32 is a point on the graph of R. Plot the point 16, 32 and graph the line y = 3 using dashes. STEP 6: The real zeros of the numerator, 0 and 1, and the real zeros of the denominator, - 4 and 3, divide the x-axis into five intervals: 1 - q , - 42 1 - 4, 02 10, 12 11, 32 13, q 2 Construct Table 12. See page 360. Plot the points from Table 12. Figure 34(a) shows the graph so far. STEP 7: Since the graph of R is above the x-axis for x 6 - 4 and only crosses the line y = 3 at 16, 32 , as x approaches - q the graph of R will approach the horizontal asymptote y = 3 from above 1 lim R 1x2 = 32. The graph xS -q of R will approach the vertical asymptote x = - 4 at the top to the left of x = - 4 1 lim -R 1x2 = q 2 and at the bottom to the right of x = - 4 x S -4 1 lim +R 1x2 = - q 2 . The graph of R will approach the vertical asymptote x S -4 360 CHAPTER 5 Polynomial and Rational Functions Table 12 –4 0 ( - q , - 4) Interval 1 3 x ( - 4, 0) (0, 1) (1, 3) (3, q ) 2 4 Number chosen -5 -2 1 2 Value of R R( - 5) = 11.25 R( - 2) = - 1.8 1 1 Ra b = 2 15 R(2) = - 1 R(4) = 4.5 Location of graph Above x-axis Below x-axis Above x-axis Below x-axis Above x-axis Point on graph ( - 5, 11.25) ( - 2, - 1.8) 1 1 a , b 2 15 (2, - 1) (4, 4.5) x = 3 at the bottom to the left of x = 3 1 lim-R 1x2 = - q 2 and at the top to xS3 the right of x = 3 1 lim+R 1x2 = q 2 . S x 3 We do not know whether the graph of R crosses or touches the line y = 3 at 16, 32. To see whether the graph, in fact, crosses or touches the line y = 3, plot 63 an additional point to the right of 16, 32. We use x = 7 to find R 172 = 6 3. 22 The graph crosses y = 3 at x = 6. Because 16, 32 is the only point where the graph of R intersects the asymptote y = 3, the graph must approach the line y = 3 from below as x S q 1 lim R 1x2 = 32 . x Sq The graph crosses the x-axis at x = 0, changing from being below the x-axis to being above. The graph also crosses the x-axis at x = 1, changing from being above the x-axis to being below. See Figure 34(b). The complete graph is shown in Figure 34(c). x = −4 (−5, 11.25) y x 4 x=3 (5, 11.25) 10 1 ( 1–2 , –– 15) (4, 4.5) (6, 3) (0, 0) −5 (−2, −1.8) y (2, −1) 10 1 ( 1–2 , –– 15) y=3 (4, 4.5) (6, 3) (0, 0) 5 x 5 x3 (2, 1.8) (2, 1) 5 y3 x –– ) (7, 63 22 (1, 0) (1, 0) −10 – 10 (b) (a) x 4 (5, 11.25) y x3 10 1 ( 1–2 , –– 15) (4, 4.5) (6, 3) (0, 0) 5 (2, 1.8) (2, 1) 5 y3 x –– ) (7, 63 22 (1, 0) – 10 Figure 34 (c) r SECTION 5.3 The Graph of a Rational Function 361 Exploration 3x2 - 3x x + x - 12 Result Figure 35(a) shows the graph. The graph does not clearly display the behavior of the function between the two x-intercepts, 0 and 1. Nor does it clearly display the fact that the graph crosses the horizontal asymptote at (6, 3). To see these parts better, graph R for - 1 … x … 2 [Figure 35(b)] and for 4 … x … 60 [Figure 35(c)]. Graph the rational function: R(x) = 2 3.5 0.5 10 21 210 2 y5 3 10 Figure 35 4 21 210 (a) 60 2.5 (b) (c) The new graphs reflect the behavior produced by the analysis. Furthermore, we observe two turning points, one between 0 and 1 and the other to the right of 6. Rounded to two decimal places, these turning points are (0.52, 0.07) and (11.48, 2.75). Now Work EXAMPL E 5 PROBLEM 31 Analyzing the Graph of a Rational Function with a Hole Analyze the graph of the rational function: R 1x2 = Solution 2x2 - 5x + 2 x2 - 4 STEP 1: Factor R and obtain R 1x2 = 12x - 12 1x - 22 1x + 22 1x - 22 The domain of R is 5 x x ≠ - 2, x ≠ 26 . STEP 2: In lowest terms, R 1x2 = 2x - 1 x + 2 x ≠ - 2, x ≠ 2 1 1 STEP 3: The y-intercept is R 102 = - . Plot the point a0, - b . The graph has one 2 2 1 1 x-intercept, , with odd multiplicity. Plot the point ¢ , 0≤. The graph will 2 2 1 cross the x-axis at x = . See Figure 36(a) on page 363. 2 STEP 4: Since x + 2 is the only factor of the denominator of R 1x2 in lowest terms, the graph has one vertical asymptote, x = - 2. However, the rational function is undefined at both x = 2 and x = - 2. Graph the line x = - 2 using dashes. The multiplicity of - 2 is odd, so the graph will approach q on one side of the vertical asymptote and - q on the other side. STEP 5: Since the degree of the numerator equals the degree of the denominator, the graph has a horizontal asymptote. To find it, form the quotient of the leading coefficient of the numerator, 2, and the leading coefficient of the denominator, 1. The graph of R has the horizontal asymptote y = 2. Graph the line y = 2 using dashes. 362 CHAPTER 5 Polynomial and Rational Functions To find out whether the graph of R intersects the horizontal asymptote y = 2, solve the equation R 1x2 = 2. R 1x2 = 2x - 1 x + 2 2x - 1 2x - 1 -1 = 2 = 21x + 22 = 2x + 4 = 4 Impossible The graph does not intersect the line y = 2. STEP 6: Look at the factored expression for R in Step 1. The real zeros of the numerator 1 and denominator, - 2, , and 2, divide the x-axis into four intervals: 2 1 - q , - 22 1 a - 2, b 2 1 a , 2b 2 12, q 2 Construct Table 13. Plot the points in Table 13. –2 Table 13 1/2 2 x Interval ( - q , - 2) 1 a - 2, b 2 1 a , 2b 2 (2, q ) Number chosen -3 -1 1 3 R(3) = 1 Value of R R( - 3) = 7 R( - 1) = - 3 1 R(1) = 3 Location of graph Above x-axis Below x-axis Above x-axis Above x-axis Point on graph ( - 3, 7) ( - 1, - 3) 1 a1, b 3 (3, 1) NOTE The coordinates of the hole were obtained by evaluating R in lowest terms 2x - 1 , at x = 2. R in lowest terms is x + 2 2(2) - 1 3 which, at x = 2, is = . ■ 2 + 2 4 STEP 7: From Table 13 we know that the graph of R is above the x-axis for x 6 - 2. From Step 5 we know that the graph of R does not intersect the asymptote y = 2. Therefore, the graph of R will approach y = 2 from above as x S - q and will approach the vertical asymptote x = - 2 at the top from the left. 1 Since the graph of R is below the x-axis for - 2 6 x 6 , the graph will 2 approach x = - 2 at the bottom from the right. Finally, since the graph of 1 R is above the x-axis for x 7 and does not intersect the horizontal asymptote 2 y = 2, the graph of R will approach y = 2 from below as x S q . The graph 1 crosses the x-axis at x = , changing from being below the x-axis to being 2 above. See Figure 36(a). See Figure 36(b) for the complete graph. Since R is not defined at 2, 3 there is a hole at the point a2, b . 4 Exploration 2x2 - 5x + 2 3 . Do you see the hole at a2, b? TRACE along the graph. Did you obtain an 4 x2 - 4 ERROR at x = 2? Are you convinced that an algebraic analysis of a rational function is required in order to accurately interpret the graph obtained with a graphing utility? Graph R(x) = SECTION 5.3 The Graph of a Rational Function y x 2 y x 2 8 (3, 7) (3, 7) 6 4 (1, ) (0, 4 3 2 1– 3 2 ) 1 ( y2 (3, 1) 1 2 8 6 4 1–2 1– , 2 2 3 (0, 1–2 ) 4 x 3 2 0) Hole at 3 (1, ) (2, –4 ) 1– 3 2 1 2 (1, 3) (a) (b) ( y2 (3, 1) 1 (1, 3) Figure 36 363 1– , 2 2 3 x 0) r As Example 5 shows, the zeros of the denominator of a rational function give rise to either vertical asymptotes or holes on the graph. Now Work EXAMPL E 6 PROBLEM 33 Constructing a Rational Function from Its Graph Find a rational function that might have the graph shown in Figure 37. x 5 y x2 10 5 y2 15 10 5 5 10 15 x 5 10 Figure 37 Solution p 1x2 in lowest terms determines the q 1x2 x-intercepts of its graph. The graph shown in Figure 37 has x-intercepts - 2 (even multiplicity; graph touches the x-axis) and 5 (odd multiplicity; graph crosses the x-axis). So one possibility for the numerator is p 1x2 = 1x + 22 2 1x - 52. The denominator of a rational function in lowest terms determines the vertical asymptotes of its graph. The vertical asymptotes of the graph are x = - 5 and x = 2. Since R 1x2 approaches q to the left of x = - 5 and R 1x2 approaches - q to the right of x = - 5, we know that (x + 5) is a factor of odd multiplicity in q 1x2 . Also, R 1x2 approaches - q on both sides of x = 2, so (x - 2) is a factor of even multiplicity in q 1x2 . A possibility for the denominator is q 1x2 = 1x + 52 1x - 22 2. 1x + 22 2 1x - 52 So far we have R 1x2 = . 1x + 52 1x - 22 2 The numerator of a rational function R 1x2 = 364 CHAPTER 5 Polynomial and Rational Functions However, the horizontal asymptote of the graph given in Figure 37 is y = 2, so we know that the degree of the numerator must equal the degree of the denominator 2 and that the quotient of leading coefficients must be . This leads to 1 5 215 21x + 22 2 1x - 52 10 R 1x2 = 1x + 52 1x - 22 2 25 r Check: Figure 38 shows the graph of R on a graphing utility. Since Figure 38 looks similar to Figure 37, we have found a rational function R for the graph in Figure 37. Figure 38 Now Work PROBLEM 51 2 Solve Applied Problems Involving Rational Functions EX AMPLE 7 Finding the Least Cost of a Can Reynolds Metal Company manufactures aluminum cans in the shape of a cylinder 1 with a capacity of 500 cubic centimeters a literb . The top and bottom of the can are 2 made of a special aluminum alloy that costs 0.05. per square centimeter. The sides of the can are made of material that costs 0.02. per square centimeter. (a) (b) (c) (d) Solution Top r Area ⫽ r 2 r h h Lateral Surface Area ⫽ 2rh Express the cost of material for the can as a function of the radius r of the can. Use a graphing utility to graph the function C = C 1r2. What value of r will result in the least cost? What is this least cost? (a) Figure 39 illustrates the components of a can in the shape of a right circular cylinder. Notice that the material required to produce a cylindrical can of height h and radius r consists of a rectangle of area 2prh and two circles, each of area pr 2. The total cost C (in cents) of manufacturing the can is therefore C = Cost of the top and bottom + Cost of the side = 21pr 2 2 # 10.052 + 12prh2 # 10.022 3 Area ⫽ r 2 Total area of top and bottom 3 3 3 Cost/unit area Total area of side Cost/unit area = 0.10pr 2 + 0.04prh Bottom Figure 39 There is an additional restriction that the height h and radius r must be chosen so that the volume V of the can is 500 cubic centimeters. Since V = pr 2h, we have 500 = pr 2h so h = 500 pr 2 Substituting this expression for h, we find that the cost C, in cents, as a function of the radius r is C 1r2 = 0.10pr 2 + 0.04pr 60 0 0 Figure 40 10 # 500 0.10pr 3 + 20 20 = = 0.10pr 2 + 2 r r pr (b) See Figure 40 for the graph of C = C 1r2. (c) Using the MINIMUM command, the cost is least for a radius of about 3.17 centimeters. (d) The least cost is C 13.172 ≈ 9.47.. r Now Work PROBLEM 61 SECTION 5.3 The Graph of a Rational Function 365 5.3 Assess Your Understanding ‘Are You Prepared?’ The answer is given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Find the intercepts of the graph of the equation y = x2 - 1 . (pp. 159–160) x2 - 4 Concepts and Vocabulary 2. True or False Every rational function has at least one asymptote. 3. Which type of asymptote will never intersect the graph of a rational function? (a) horizontal (b) oblique (c) vertical (d) all of these 4. Identify the y-intercept of the graph of 61x - 12 R 1x2 = . 1x + 12 1x + 22 (a) - 3 (b) - 2 (c) - 1 5. R 1x2 = x1x - 22 2 x - 2 (a) Find the domain of R. (b) Find the x-intercepts of R. 6. True or False The graph of a rational function sometimes has a hole. (d) 1 Skill Building In Problems 7–50, follow Steps 1 through 7 on page 356 to analyze the graph of each function. 7. R 1x2 = x + 1 x1x + 42 8. R 1x2 = x 1x - 12 1x + 22 9. R 1x2 = 3x + 3 2x + 4 10. R 1x2 = 2x + 4 x - 1 11. R 1x2 = 3 x - 4 12. R 1x2 = 6 x - x - 6 13. P 1x2 = x4 + x2 + 1 x2 - 1 14. Q1x2 = x4 - 1 x2 - 4 15. H1x2 = x3 - 1 x2 - 9 16. G1x2 = x3 + 1 x2 + 2x 17. R 1x2 = x2 x2 + x - 6 18. R 1x2 = x2 + x - 12 x2 - 4 19. G1x2 = x x - 4 20. G1x2 = 3x x - 1 21. R 1x2 = 3 1x - 12 1x2 - 42 22. R 1x2 = -4 1x + 12 1x2 - 92 23. H1x2 = x2 - 1 x4 - 16 24. H1x2 = x2 + 4 x4 - 1 25. F 1x2 = x2 - 3x - 4 x + 2 26. F 1x2 = x2 + 3x + 2 x - 1 27. R 1x2 = x2 + x - 12 x - 4 28. R 1x2 = x2 - x - 12 x + 5 29. F 1x2 = x2 + x - 12 x + 2 30. G1x2 = x2 - x - 12 x + 1 31. R 1x2 = x1x - 12 2 34. R 1x2 = x2 + 3x - 10 x2 + 8x + 15 35. R 1x2 = 37. R 1x2 = x2 + 5x + 6 x + 3 38. R 1x2 = 40. H1x2 = 2 - 2x x2 - 1 41. F 1x2 = 44. G1x2 = 2 - x 1x - 12 2 45. f 1x2 = x + 1 x 46. f 1x2 = 2x + 9 x 49. f 1x2 = x + 1 x3 50. f 1x2 = 2x + 9 x3 2 2 16 x 2 32. R 1x2 = 1x + 32 3 48. f 1x2 = 2x2 + 2 x2 - 5x + 4 x2 - 2x + 1 1x - 12 1x + 22 1x - 32 33. R 1x2 = x2 + x - 12 x2 - x - 6 6x2 - 7x - 3 2x2 - 7x + 6 36. R 1x2 = 8x2 + 26x + 15 2x2 - x - 15 x2 + x - 30 x + 6 39. H1x2 = 3x - 6 4 - x2 x1x - 42 2 42. F 1x2 = x2 - 2x - 15 x2 + 6x + 9 43. G1x2 = x 1x + 22 2 47. f 1x2 = x2 + 1 x 366 CHAPTER 5 Polynomial and Rational Functions In Problems 51–54, find a rational function that might have the given graph. (More than one answer might be possible.) 51. x 2 52. x2 y x 1 y 3 x1 3 y1 y0 3 3 x 3 3 53. x 3 3 54. y x 3 y 10 3 x4 8 2 y1 6 4 4 3 2 1 1 3 4 5 x 2 x 1 y3 2 15 10 5 x2 5 10 15 20 x 2 4 6 8 Applications and Extensions 55. Drug Concentration The concentration C of a certain drug in a patient’s bloodstream t hours after injection is given by C 1t2 = t 2t 2 + 1 (a) Find the horizontal asymptote of C 1t2 . What happens to the concentration of the drug as t increases? (b) Using your graphing utility, graph C = C 1t2. (c) Determine the time at which the concentration is highest. 56. Drug Concentration The concentration C of a certain drug in a patient’s bloodstream t minutes after injection is given by 50t C 1t2 = 2 t + 25 (a) Find the horizontal asymptote of C 1t2 . What happens to the concentration of the drug as t increases? (b) Using your graphing utility, graph C = C 1t2. (c) Determine the time at which the concentration is highest. 57. Minimum Cost A rectangular area adjacent to a river is to be fenced in; no fence is needed on the river side. The enclosed area is to be 1000 square feet. Fencing for the side parallel to the river is $5 per linear foot, and fencing for the other two sides is $8 per linear foot; the four corner posts are $25 apiece. Let x be the length of one of the sides perpendicular to the river. (a) Write a function C 1x2 that describes the cost of the project. (b) What is the domain of C? (c) Use a graphing utility to graph C = C 1x2. (d) Find the dimensions of the cheapest enclosure. Source: http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/ pandr/rational/rational.html 58. Doppler Effect The Doppler effect (named after Christian Doppler) is the change in the pitch (frequency) of the sound from a source (s) as heard by an observer (o) when one or both are in motion. If we assume both the source and the observer are moving in the same direction, the relationship is f′ = fa a where f′ fa v vo vs = = = = = v - vo b v - vs perceived pitch by the observer actual pitch of the source speed of sound in air (assume 772.4 mph) speed of the observer speed of the source Suppose that you are traveling down the road at 45 mph and you hear an ambulance (with siren) coming toward you from the rear. The actual pitch of the siren is 600 hertz (Hz). (a) Write a function f′(vs) that describes this scenario. (b) If f′ = 620 Hz, find the speed of the ambulance. (c) Use a graphing utility to graph the function. (d) Verify your answer from part (b). Source: www.acs.psu.edu/drussell/ SECTION 5.3 The Graph of a Rational Function 59. Minimizing Surface Area United Parcel Service has contracted you to design a closed box with a square base that has a volume of 10,000 cubic inches. See the illustration. (a) Express the surface area S of the box as a function of x. y (b) Using a graphing utility, graph the function found in part (a). x (c) What is the minimum amount of x cardboard that can be used to construct the box? (d) What are the dimensions of the box that minimize the surface area? (e) Why might UPS be interested in designing a box that minimizes the surface area? 60. Minimizing Surface Area United Parcel y Service has contracted you to design an open box with a square base that has a volume of x 5000 cubic inches. See the illustration. x (a) Express the surface area S of the box as a function of x. (b) Using a graphing utility, graph the function found in part (a). (c) What is the minimum amount of cardboard that can be used to construct the box? (d) What are the dimensions of the box that minimize the surface area? (e) Why might UPS be interested in designing a box that minimizes the surface area? 61. Cost of a Can A can in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top 367 and bottom are made of material that costs 6¢ per square centimeter, while the sides are made of material that costs 4¢ per square centimeter. (a) Express the total cost C of the material as a function of the radius r of the cylinder. (Refer to Figure 39.) (b) Graph C = C 1r2. For what value of r is the cost C a minimum? 62. Material Needed to Make a Drum A steel drum in the shape of a right circular cylinder is required to have a volume of 100 cubic feet. (a) Express the amount A of material required to make the drum as a function of the radius r of the cylinder. (b) How much material is required if the drum’s radius is 3 feet? (c) How much material is required if the drum’s radius is 4 feet? (d) How much material is required if the drum’s radius is 5 feet? (e) Graph A = A(r). For what value of r is A smallest? Discussion and Writing at - 1; one vertical asymptote at x = - 5 and another at x = 6; and one horizontal asymptote, y = 3. Compare your function to a fellow classmate’s. How do they differ? What are their similarities? 63. Graph each of the following functions: 2 3 y = x - 1 x - 1 y = x - 1 x - 1 y = x4 - 1 x - 1 y = x5 - 1 x - 1 Is x = 1 a vertical asymptote? Why not? What is happening for xn - 1 ,n Ú 1 x = 1? What do you conjecture about y = x - 1 an integer, for x = 1? 67. Create a rational function that has the following characteristics: crosses the x-axis at 3; touches the x-axis at - 2; one vertical asymptote, x = 1; and one horizontal asymptote, y = 2. Give your rational function to a fellow classmate and ask for a written critique of your rational function. 65. Write a few paragraphs that provide a general strategy for graphing a rational function. Be sure to mention the following: proper, improper, intercepts, and asymptotes. 68. Create a rational function with the following characteristics: three real zeros, one of multiplicity 2; y-intercept 1; vertical asymptotes, x = - 2 and x = 3; oblique asymptote, y = 2x + 1. Is this rational function unique? Compare your function with those of other students. What will be the same as everyone else’s? Add some more characteristics, such as symmetry or naming the real zeros. How does this modify the rational function? 66. Create a rational function that has the following characteristics: crosses the x-axis at 2; touches the x-axis 69. Explain the circumstances under which the graph of a rational function will have a hole. 64. Graph each of the following functions: y = x2 x - 1 y = x4 x - 1 y = x6 x - 1 y = x8 x - 1 What similarities do you see? What differences? Retain Your Knowledge Problems 70–73 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 70. Subtract: (4x3 - 7x + 1) - (5x2 - 9x + 3) 71. Solve: x - 2 3x = 3x + 1 x + 5 2 72. Find the maximum value of f 1x2 = - x2 + 6x - 5. 3 25 - 3 . Round your answer to three 73. Approximate 27 + 2 decimal places. 368 CHAPTER 5 Polynomial and Rational Functions ‘Are You Prepared?’ Answers 1 1. ¢0, ≤, 11, 02, 1 - 1, 02 4 5.4 Polynomial and Rational Inequalities PREPARING FOR THIS SECTION Before getting started, review the following: r Solving Linear Inequalities (Section 1.5, pp. 123–125) r Solving Quadratic Inequalities (Section 4.5, pp. 312–313) Now Work the ‘Are You Prepared?’ problems on page 372. OBJECTIVES 1 Solve Polynomial Inequalities (p. 368) 2 Solve Rational Inequalities (p. 370) 1 Solve Polynomial Inequalities In this section we solve inequalities that involve polynomials of degree 3 and higher, along with inequalities that involve rational functions. To help understand the algebraic procedure for solving such inequalities, we use the information obtained in the previous three sections about the graphs of polynomial and rational functions. The approach follows the same methodology that we used to solve inequalities involving quadratic functions. EX AMPLE 1 Solution Solving a Polynomial Inequality Using Its Graph Solve 1x + 32 1x - 12 2 7 0 by graphing f1x2 = 1x + 32 1x - 12 2. Graph f 1x2 = 1x + 32 1x - 12 2 and determine the intervals of x for which the graph is above the x-axis. These values of x result in f 1x2 being positive. Using Steps 1 through 5 on page 335, we obtain the graph shown in Figure 41. y 12 End behavior y = x3 (22, 9) 9 6 3 (23, 0) 24 22 3 End behavior 3 y=x 2 (1, 0) 4 x 6 Figure 41 f (x) = (x + 3)(x - 1)2 From the graph, we can see that f 1x2 7 0 for - 3 6 x 6 1 or x 7 1. The solution set is 5 x - 3 6 x 6 1 or x 7 16 or, using interval notation, 1 - 3, 12 h 11, q 2 . r Now Work PROBLEM 9 The results of Example 1 lead to the following approach to solving polynomial and rational inequalities algebraically. Suppose that the polynomial or rational inequality is in one of the forms f1x2 6 0 f1x2 7 0 f1x2 … 0 f1x2 Ú 0 369 SECTION 5.4 Polynomial and Rational Inequalities Locate the zeros of f if f is a polynomial function, and locate the zeros of the numerator and the denominator if f is a rational function. If we use these zeros to divide the real number line into intervals, we know that on each interval, the graph of f is either above the x-axis 3 f1x2 7 04 or below the x-axis 3 f 1x2 6 04 . This will enable us to identify the solution of the inequality. EXAMPL E 2 How to Solve a Polynomial Inequality Algebraically Solve the inequality x4 7 x algebraically, and graph the solution set. Step-by-Step Solution Rearrange the inequality so that 0 is on the right side. Step 1: Write the inequality so that a polynomial expression f is on the left side and zero is on the right side. x4 7 x x4 - x 7 0 Subtract x from both sides of the inequality. This inequality is equivalent to the one we wish to solve. Step 2: Determine the real zeros (x-intercepts of the graph) of f. Find the real zeros of f1x2 = x4 - x by solving x4 - x = 0. x = 0 or x = 0 or x4 - x x1x3 - 12 2 x1x - 12 1x + x + 12 x - 1 = 0 or x2 + x + 1 x = 1 = = = = 0 0 Factor out x. 0 Factor the difference of two cubes. 0 Set each factor equal to zero and solve. The equation x2 + x + 1 = 0 has no real solutions. Do you see why? Step 3: Use the zeros found in Step 2 to divide the real number line into intervals. Use the real zeros to separate the real number line into three intervals: Step 4: Select a number in each interval, evaluate f at the number, and determine whether f (x) is positive or negative. If f (x) is positive, all values of f in the interval are positive. If f (x) is negative, all values of f in the interval are negative. Select a test number in each interval found in Step 3 and evaluate f1x2 = x4 - x at each number to determine whether f 1x2 is positive or negative. See Table 14. NOTE If the inequality is not strict (that is, if it is … or Ú ), include the solutions of f (x) = 0 in the solution set. ■ –2 –1 0 1 2 Figure 42 x 1 - q , 02 10, 12 Table 14 11, q 2 0 1 x Interval ( - q , 0) (0, 1) (1, q ) Number chosen -1 1 2 2 Value of f f ( - 1) = 2 1 7 fa b = 2 16 f (2) = 14 Conclusion Positive Negative Positive Since we want to know where f1x2 is positive, conclude that f1x2 7 0 for all numbers x for which x 6 0 or x 7 1. Because the original inequality is strict, numbers x that satisfy the equation x4 = x are not solutions. The solution set of the inequality x4 7 x is {x x 6 0 or x 7 1} or, using interval notation, 1 - q , 02 h 11, q 2 . Figure 42 shows the graph of the solution set. r The Role of Multiplicity in Solving Polynomial Inequalities In Example 2, we used the number - 1 and found that f is positive for all x 6 0. Because the “cut point” of 0 is the result of a zero of odd multiplicity (x is a factor to the first power), we know that the sign of f will change on either side of 0, so 370 CHAPTER 5 Polynomial and Rational Functions for 0 6 x 6 1, f(x) will be negative. Similarly, we know that f(x) will be positive for x 7 1, since the multiplicity of the zero 1 is odd. Therefore, the solution set of x4 7 x is {x x 6 0 or x 7 16 or, using interval notation, 1 - q , 02 h 11, q 2. Now Work PROBLEM 21 2 Solve Rational Inequalities Just as we presented a graphical approach to help us understand the algebraic procedure for solving inequalities involving polynomials, we present a graphical approach to help us understand the algebraic procedure for solving inequalities involving rational expressions. EX AMPLE 3 Solving a Rational Inequality Using Its Graph Solve Solution x - 1 x - 1 Ú 0 by graphing R 1x2 = 2 . x2 - 4 x - 4 Graph R 1x2 = x - 1 and determine the intervals of x such that the graph is above x2 - 4 or on the x-axis. These values of x result in R 1x2 being positive or zero. We graphed x - 1 R 1x2 = 2 in Example 1, Section 5.3 (pp. 353–355). We reproduce the graph in x - 4 Figure 43. x 2 x y 2 3 (1, 0) 3 ( 3, 0.8) (0, 1–4) (3–2, 2–7) (3, 0.4) 3 x y 0 3 Figure 43 R 1x2 = x - 1 x2 - 4 From the graph, we can see that R 1x2 Ú 0 for - 2 6 x … 1 or x 7 2. The solution set is 5 x - 2 6 x … 1 or x 7 26 or, using interval notation, 1 - 2, 14 h 12, q 2 . r Now Work PROBLEM 15 To solve a rational inequality algebraically, we follow the same approach that we used to solve a polynomial inequality algebraically. However, we must also identify the zeros of the denominator of the rational function, because the sign of a rational function may change on either side of a vertical asymptote. Convince yourself of this by looking at Figure 43. Notice that the function values are negative for x 6 - 2 and are positive for x 7 - 2 (but less than 1). EX AMPLE 4 How to Solve a Rational Inequality Algebraically Solve the inequality 3x2 + 13x + 9 … 3 algebraically, and graph the solution set. 1x + 22 2 SECTION 5.4 Polynomial and Rational Inequalities Step-by-Step Solution 371 Rearrange the inequality so that 0 is on the right side. 3x2 + 13x + 9 … 3 1x + 22 2 Step 1: Write the inequality so that a rational expression f is on the left side and zero is on the right side. 3x2 + 13x + 9 - 3 … 0 x2 + 4x + 4 Subtract 3 from both sides of the inequality; Expand (x + 2)2. 3x2 + 13x + 9 x2 + 4x + 4 # 3 … 0 x2 + 4x + 4 x2 + 4x + 4 Multiply 3 by 3x2 + 13x + 9 - 3x2 - 12x - 12 … 0 x2 + 4x + 4 Write as a single quotient. x - 3 … 0 1x + 22 2 x 2 + 4x + 4 . x 2 + 4x + 4 Combine like terms. x - 3 is 3. Also, f is undefined for x = - 2. 1x + 22 2 Step 2: Determine the real zeros (x-intercepts of the graph) of f and the real numbers for which f is undefined. The zero of f1x2 = Step 3: Use the zeros and undefined values found in Step 2 to divide the real number line into intervals. Use the zero and the undefined value to separate the real number line into three intervals: Step 4: Select a number in each interval, evaluate f at the number, and determine whether f (x) is positive or negative. If f (x) is positive, all values of f in the interval are positive. If f (x) is negative, all values of f in the interval are negative. Select a test number in each interval from Step 3, and evaluate f at each number to determine whether f 1x 2 is positive or negative. See Table 15. NOTE If the inequality is not strict ( … or Ú ), include the solutions of f(x) = 0 in the solution set. ■ 6 4 Figure 44 2 0 2 4 6 x 1 - q , - 22 1 - 2, 32 13, q 2 –2 Table 15 3 x Interval ( - q , - 2) ( - 2, 3) (3, q ) Number chosen -3 0 4 Value of f f( - 3) = - 6 f (0) = - Conclusion Negative Negative 3 4 f (4) = 1 36 Positive Since we want to know where f(x) is negative or zero, we conclude that f (x) … 0 for all numbers for which x 6 - 2 or - 2 6 x … 3. Notice that we do not include - 2 in the solution because - 2 is not in the domain of f. The solution set of the 3x2 + 13x + 9 inequality … 3 is {x x 6 - 2 or - 2 6 x … 3} or, using interval 1x + 22 2 notation, 1 - q , - 22 h 1 - 2, 3]. Figure 44 shows the graph of the solution set. r The Role of Multiplicity in Solving Rational Inequalities In Example 4, we used the number - 3 and found that f(x) is negative for all x 6 - 2. Because the “cut point” of - 2 is the result of a zero of even multiplicity, we know the sign of f(x) will not change on either side of - 2, so for - 2 6 x 6 3, f(x) will also be negative. Because the “cut point” of 3 is the result of a zero of odd multiplicity, the sign of f(x) will change on either side of 3, so for x 7 3, f(x) will be positive. 3x2 + 13x + 9 Therefore, the solution set of … 3 is {x x 6 - 2 or - 2 6 x … 3} or, 1x + 22 2 using interval notation, 1 - q , - 22 h 1 - 2, 3]. Now Work PROBLEMS 33 and 39 372 CHAPTER 5 Polynomial and Rational Functions SUMMARY Steps for Solving Polynomial and Rational Inequalities Algebraically STEP 1: Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right side in one of the following forms: f 1x2 7 0 f1x2 Ú 0 f1x2 6 0 f1x2 … 0 For rational expressions, be sure that the left side is written as a single quotient, and find the domain of f . STEP 2: Determine the real numbers at which the expression f equals zero and, if the expression is rational, the real numbers at which the expression f is undefined. STEP 3: Use the numbers found in Step 2 to separate the real number line into intervals. STEP 4: Select a number in each interval and evaluate f at the number. (a) If the value of f is positive, then f1x2 7 0 for all numbers x in the interval. (b) If the value of f is negative, then f1x2 6 0 for all numbers x in the interval. If the inequality is not strict 1 Ú or … 2 , include the solutions of f1x2 = 0 that are in the domain of f in the solution set. Be careful to exclude values of x where f is undefined. 5.4 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Solve the inequality 3 - 4x 7 5. Graph the solution set. (pp. 123–125) 2. Solve the inequality x2 - 5x … 24. Graph the solution set. (pp. 312–313) Concepts and Vocabulary 3. Which of the following could be a test number for the interval - 2 6 x 6 5? (a) - 3 (b) - 2 (c) 4 (d) 7 4. True or False The graph of f 1x2 = x is above the x-axis x - 3 for x 6 0 or x 7 3, so the solution set of the inequality x Ú 0 is 5x x … 0 or x Ú 36 . x - 3 Skill Building In Problems 5–8, use the graph of the function f to solve the inequality. 5. (a) f 1x2 7 0 (b) f 1x2 … 0 6. (a) f 1x2 6 0 (b) f 1x2 Ú 0 y 2 y 1 0 1 2 –2 x –1 1 –1 –2 7. (a) f 1x2 6 0 (b) f 1x2 Ú 0 x 1 y 8. (a) f 1x2 7 0 (b) f 1x2 … 0 x1 3 y 3 2 y0 3 3 4 3 2 3 y1 x 1 1 3 2 x 1 x2 4 5 x 2 3 x SECTION 5.4 Polynomial and Rational Inequalities 373 In Problems 9–14, solve the inequality by using the graph of the function. [Hint: The graphs were drawn in Problems 81–86 of Section 5.1.] 9. Solve f 1x2 6 0, where f 1x2 = x2 1x - 32 . 10. Solve f 1x2 … 0, where f 1x2 = x1x + 22 2. 11. Solve f 1x2 Ú 0, where f 1x2 = 1x + 42 2 11 - x2 . 12. Solve f 1x2 7 0, where f 1x2 = 1x - 12 1x + 32 2. 13. Solve f 1x2 … 0, where f 1x2 = - 21x + 22 1x - 22 3. 1 14. Solve f 1x2 6 0, where f 1x2 = - 1x + 42 1x - 12 3. 2 In Problems 15–18, solve the inequality by using the graph of the function. [Hint: The graphs were drawn in Problems 7–10 of Section 5.3.] 15. Solve R 1x2 7 0, where R 1x2 = x + 1 . x1x + 42 16. Solve R 1x2 6 0, where R 1x2 = x . 1x - 12 1x + 22 17. Solve R 1x2 … 0, where R 1x2 = 3x + 3 . 2x + 4 18. Solve R 1x2 Ú 0, where R 1x2 = 2x + 4 . x - 1 In Problems 19–48, solve each inequality algebraically. 19. 1x - 52 2 1x + 22 6 0 20. 1x - 52 1x + 22 2 7 0 21. x3 - 4x2 7 0 22. x3 + 8x2 6 0 23. 2x3 7 - 8x2 24. 3x3 6 - 15x2 25. 1x - 12 1x - 22 1x - 32 … 0 26. 1x + 12 1x + 22 1x + 32 … 0 27. x3 - 2x2 - 3x 7 0 28. x3 + 2x2 - 3x 7 0 29. x4 7 x2 30. x4 6 9x2 31. x4 7 1 32. x3 7 1 33. 34. 37. x - 3 7 0 x + 1 1x - 22 2 x2 - 1 35. 38. Ú 0 1x - 12 1x + 12 x 1x + 52 2 x2 - 4 36. … 0 Ú 0 x - 4 Ú 1 2x + 4 41. 3x - 5 … 2 x + 2 42. 43. 2 1 6 x - 2 3x - 9 44. 3 5 7 x - 3 x + 1 45. 1x - 12 1x + 12 47. Ú 0 13 - x2 3 12x + 12 x3 - 1 6 0 x - 1 x + 4 … 1 x - 2 x + 2 Ú 1 x - 4 x1x2 + 12 1x - 22 1x - 32 1x + 22 39. 40. 46. x + 1 7 0 x - 1 48. … 0 x2 13 + x2 1x + 42 1x + 52 1x - 12 12 - x2 3 13x - 22 x3 + 1 Ú 0 6 0 Mixed Practice In Problems 49–60, solve each inequality algebraically. 49. 1x + 12 1x - 32 1x - 52 7 0 50. 12x - 12 1x + 22 1x + 52 6 0 52. x2 + 3x Ú 10 53. 55. 31x2 - 22 6 21x - 12 2 + x2 56. 1x - 32 1x + 22 6 x2 + 3x + 5 57. 6x - 5 6 59. x3 - 9x … 0 60. x3 - x Ú 0 58. x + 12 6 7 x x + 1 … 2 x - 3 51. 7x - 4 Ú - 2x2 54. x - 1 Ú -2 x + 2 6 x In Problems 61–64, (a) graph each function by hand, and (b) solve f 1x2 Ú 0. 61. f 1x2 = x2 + 5x - 6 x2 - 4x + 4 62. f 1x2 = 2x2 + 9x + 9 x2 - 4 63. f 1x2 = x3 + 2x2 - 11x - 12 x2 - x - 6 64. f 1x2 = x3 - 6x2 + 9x - 4 x2 + x - 20 374 CHAPTER 5 Polynomial and Rational Functions Applications and Extensions 65. For what positive numbers will the cube of a number exceed four times its square? 66. For what positive numbers will the cube of a number be less than the number? 67. What is the domain of the function f 1x2 = 2x4 - 16? 68. What is the domain of the function f 1x2 = 2x3 - 3x2? 69. What is the domain of the function f 1x2 = x Ax x 70. What is the domain of the function f 1x2 = Ax + + 2 ? 4 1 ? 4 In Problems 71–74, determine where the graph of f is below the graph of g by solving the inequality f 1x2 … g1x2. Graph f and g together. 71. f 1x2 = x4 - 1 g1x2 = - 2x2 + 2 73. f 1x2 = x - 4 4 g1x2 = 3x2 72. f 1x2 = x4 - 1 g1x2 = x - 1 K = 2W 1S + L2 S2 where W = weight of the jumper (pounds) K = cord’s stiffness (pounds per foot) L = free length of the cord (feet) S = stretch (feet) (a) A 150-pound person plans to jump off a ledge attached to a cord of length 42 feet. If the stiffness of the cord is no less than 16 pounds per foot, how much will the cord stretch? (b) If safety requirements will not permit the jumper to get any closer than 3 feet to the ground, what is the minimum height required for the ledge in part (a)? Source: American Institute of Physics, Physics News Update, No. 150, November 5, 1993. 78. Gravitational Force According to Newton’s Law of Universal Gravitation, the attractive force F between two bodies is given by 74. f 1x2 = x4 g1x2 = 2 - x2 75. Average Cost Suppose that the daily cost C of manufacturing bicycles is given by C 1x2 = 80x + 5000. Then the average 80x + 5000 daily cost C is given by C 1x2 = . How many x bicycles must be produced each day for the average cost to be no more than $100? 76. Average Cost See Problem 75. Suppose that the government imposes a $1000-per-day tax on the bicycle manufacturer so that the daily cost C of manufacturing x bicycles is now given by C 1x2 = 80x + 6000. Now the average daily cost C is 80x + 6000 given by C 1x2 = . How many bicycles must be x produced each day for the average cost to be no more than $100? 77. Bungee Jumping Originating on Pentecost Island in the Pacific, the practice of a person jumping from a high place harnessed to a flexible attachment was introduced to Western culture in 1979 by the Oxford University Dangerous Sport Club. One important parameter to know before attempting a bungee jump is the amount the cord will stretch at the bottom of the fall. The stiffness of the cord is related to the amount of stretch by the equation F = G m1m2 r2 where m1, m2 = the masses of the two bodies r = distance between the two bodies G = gravitational constant = 6.6742 * 10 - 11 newtons # meter 2 # kilogram-2 Suppose an object is traveling directly from Earth to the moon. The mass of Earth is 5.9742 * 1024 kilograms, the mass of the moon is 7.349 * 1022 kilograms, and the mean distance from Earth to the moon is 384,400 kilometers. For an object between Earth and the moon, how far from Earth is the force on the object due to the moon greater than the force on the object due to Earth? Source: www.solarviews.com;en.wikipedia.org 79. Field Trip Mrs. West has decided to take her fifth grade class to a play. The manager of the theater agreed to discount the regular $40 price of the ticket by $0.20 for each ticket sold. The cost of the bus, $500, will be split equally among the students. How many students must attend to keep the cost per student at or below $40? Explaining Concepts: Discussion and Writing 80. Make up an inequality that has no solution. Make up one that has exactly one solution. 81. The inequality x4 + 1 6 - 5 has no solution. Explain why. x + 4 … 0 by 82. A student attempted to solve the inequality x - 3 multiplying both sides of the inequality by x - 3 to get x + 4 … 0. This led to a solution of 5x x … - 46 . Is the student correct? Explain. 83. Write a rational {x - 3 6 x … 5}. inequality whose solution set Retain Your Knowledge Problems 84–87 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 84. Solve: 9 - 2x … 4x + 1 86. Factor completely: 6x4y4 + 3x3y5 - 18x2y6 85. Given f 1x2 = x2 + 3x - 2, find f 1x - 22. 87. Suppose y varies directly with 2x. Write a general formula to describe the variation if y = 2 when x = 9. is SECTION 5.5 The Real Zeros of a Polynomial Function 375 ‘Are You Prepared?’ Answers 1 1 1. e x ` x 6 - f or a - q , - b 2 2 2. {x - 3 … x … 8} or [ - 3, 8] 2 1 1–2 0 –3 1 0 8 5.5 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: r Evaluating Functions (Section 3.1, pp. 202–204) r Factoring Polynomials (Chapter R, Section R.5, pp. 49–55) r Synthetic Division (Chapter R, Section R.6, pp. 58–61) r Polynomial Division (Chapter R, Section R.4, pp. 44–47) r Solve a Quadratic Equation (Section 1.2, pp. 92–99) Now Work the ‘Are You Prepared?’ problems on page 386. OBJECTIVES 1 Use the Remainder and Factor Theorems (p. 375) 2 Use Descartes’ Rule of Signs to Determine the Number of Positive and the Number of Negative Real Zeros of a Polynomial Function (p. 378) 3 Use the Rational Zeros Theorem to List the Potential Rational Zeros of a Polynomial Function (p. 379) 4 Find the Real Zeros of a Polynomial Function (p. 380) 5 Solve Polynomial Equations (p. 382) 6 Use the Theorem for Bounds on Zeros (p. 383) 7 Use the Intermediate Value Theorem (p. 384) In Section 5.1, we were able to identify the real zeros of a polynomial function because either the polynomial function was in factored form or it could be easily factored. But how do we find the real zeros of a polynomial function if it is not factored or cannot be easily factored? Recall that if r is a real zero of a polynomial function f, then f1r2 = 0, r is an x-intercept of the graph of f, x - r is a factor of f, and r is a solution of the equation f1x2 = 0. For example, if x - 4 is a factor of f, then 4 is a real zero of f and 4 is a solution to the equation f1x2 = 0. For polynomial functions, we have seen the importance of the real zeros for graphing. In most cases, however, the real zeros of a polynomial function are difficult to find using algebraic methods. No nice formulas like the quadratic formula are available to help us find zeros for polynomials of degree 3 or higher. Formulas do exist for solving any third- or fourth-degree polynomial equation, but they are somewhat complicated. No general formulas exist for polynomial equations of degree 5 or higher. Refer to the Historical Feature at the end of this section for more information. 1 Use the Remainder and Factor Theorems When one polynomial (the dividend) is divided by another (the divisor), a quotient polynomial and a remainder are obtained, the remainder being either the zero polynomial or a polynomial whose degree is less than the degree of the divisor. To check, verify that 1Quotient2 1Divisor2 + Remainder = Dividend This checking routine is the basis for a famous theorem called the division algorithm* for polynomials, which we now state without proof. *A systematic process in which certain steps are repeated a finite number of times is called an algorithm. For example, long division is an algorithm. 376 CHAPTER 5 Polynomial and Rational Functions THEOREM Division Algorithm for Polynomials If f1x2 and g1x2 denote polynomial functions and if g1x2 is a polynomial whose degree is greater than zero, then there are unique polynomial functions q 1x2 and r 1x2 such that r 1x2 f1x2 = q 1x2 + g1x2 g1x2 or f1x2 = q 1x2g1x2 + r 1x2 c dividend c quotient c (1) c divisor remainder where r 1x2 is either the zero polynomial or a polynomial of degree less than that of g1x2 . In equation (1), f1x2 is the dividend, g1x2 is the divisor, q 1x2 is the quotient, and r 1x2 is the remainder. If the divisor g1x2 is a first-degree polynomial of the form g1x2 = x - c c a real number then the remainder r 1x2 is either the zero polynomial or a polynomial of degree 0. As a result, for such divisors, the remainder is some number, say R, and we may write f1x2 = 1x - c2q 1x2 + R (2) This equation is an identity in x and is true for all real numbers x. Suppose that x = c. Then equation (2) becomes f 1c2 = 1c - c2q 1c2 + R f1c2 = R Substitute f1c2 for R in equation (2) to obtain f1x2 = 1x - c2q 1x2 + f1c2 (3) which proves the Remainder Theorem. REMAINDER THEOREM EX AMPLE 1 Let f be a polynomial function. If f 1x2 is divided by x - c, then the remainder is f1c2 . Using the Remainder Theorem Find the remainder when f1x2 = x3 - 4x2 - 5 is divided by (a) x - 3 Solution (b) x + 2 (a) Either long division or synthetic division could be used, but it is easier to use the Remainder Theorem, which says that the remainder is f132 . f 132 = 132 3 - 4132 2 - 5 = 27 - 36 - 5 = - 14 The remainder is - 14. (b) To find the remainder when f1x2 is divided by x + 2 = x - ( - 2), find f( - 2). f1 - 22 = 1 - 22 3 - 41 - 22 2 - 5 = - 8 - 16 - 5 = - 29 The remainder is - 29. r Compare the method used in Example 1(a) with the method used in Example 1 of Chapter R, Section R.6. Which method do you prefer? SECTION 5.5 The Real Zeros of a Polynomial Function 377 COMMENT A graphing utility provides another way to find the value of a function using the eVALUEate feature. Consult your manual for details. Then check the results of Example 1. ■ An important and useful consequence of the Remainder Theorem is the Factor Theorem. FACTOR THEOREM Let f be a polynomial function. Then x - c is a factor of f1x2 if and only if f1c2 = 0. The Factor Theorem actually consists of two separate statements: 1. If f1c2 = 0, then x - c is a factor of f 1x2 . 2. If x - c is a factor of f1x2 , then f 1c2 = 0. The proof requires two parts. Proof 1. Suppose that f1c2 = 0. Then, by equation (3), we have f 1x2 = 1x - c2q 1x2 for some polynomial q 1x2. That is, x - c is a factor of f 1x2. 2. Suppose that x - c is a factor of f1x2. Then there is a polynomial function q such that f 1x2 = 1x - c2 q 1x2 Replacing x by c, we find that f 1c2 = 1c - c2q 1c2 = 0 # q 1c2 = 0 This completes the proof. ■ One use of the Factor Theorem is to determine whether a polynomial has a particular factor. EXAM PL E 2 Using the Factor Theorem Use the Factor Theorem to determine whether the function f1x2 = 2x3 - x2 + 2x - 3 has the factor (a) x - 1 Solution (b) x + 2 The Factor Theorem states that if f1c2 = 0, then x - c is a factor. (a) Because x - 1 is of the form x - c with c = 1, find the value of f112. We choose to use substitution. f112 = 2112 3 - 112 2 + 2112 - 3 = 2 - 1 + 2 - 3 = 0 By the Factor Theorem, x - 1 is a factor of f1x2. (b) To test the factor x + 2, first write it in the form x - c. Since x + 2 = x - 1 - 22 , find the value of f1 - 22. We choose to use synthetic division. - 2) 2 2 -1 2 - 4 10 - 5 12 -3 - 24 - 27 378 CHAPTER 5 Polynomial and Rational Functions Because f1 - 22 = - 27 ≠ 0, conclude from the Factor Theorem that x - 1 - 22 = x + 2 is not a factor of f1x2. r Now Work PROBLEM 11 In Example 2(a), x - 1 was found to be a factor of f. To write f in factored form, use long division or